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Algebra 2

Algebra 2 Practice Test: Practice Test 99

Practice Test 99 for Algebra 2: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

On the complex plane (real axis horizontal, imaginary axis vertical), the complex number z=−3+4iz = -3 + 4iz=−3+4i is represented by the point (a,b)(a,b)(a,b).

Which point corresponds to zzz, and what is ∣z∣|z|∣z∣ (the distance from the origin), where ∣a+bi∣=a2+b2|a+bi|=\sqrt{a^2+b^2}∣a+bi∣=a2+b2​?

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Question 1

On the complex plane (real axis horizontal, imaginary axis vertical), the complex number z=−3+4iz = -3 + 4iz=−3+4i is represented by the point (a,b)(a,b)(a,b).

Which point corresponds to zzz, and what is ∣z∣|z|∣z∣ (the distance from the origin), where ∣a+bi∣=a2+b2|a+bi|=\sqrt{a^2+b^2}∣a+bi∣=a2+b2​?

  1. (4,−3)(4,-3)(4,−3) and ∣z∣=5|z|=5∣z∣=5
  2. (−3,4)(-3,4)(−3,4) and ∣z∣=7|z|=7∣z∣=7
  3. (−3,4)(-3,4)(−3,4) and ∣z∣=5|z|=5∣z∣=5 (correct answer)
  4. (3,4)(3,4)(3,4) and ∣z∣=5|z|=5∣z∣=5

Explanation: This question tests your understanding of representing complex numbers geometrically on the complex plane and finding their modulus (absolute value) as the distance from the origin. The complex plane is a coordinate system where the horizontal axis represents the real part and the vertical axis represents the imaginary part: the complex number a + bi is plotted at point (a, b), just like ordered pairs! For example, 3 + 2i goes at (3, 2), and -1 - 4i goes at (-1, -4). The modulus (absolute value) of a + bi equals square root of (a squared + b squared), which is the distance from the origin to point (a, b) using the Pythagorean theorem—the real and imaginary parts form legs of right triangle, modulus is hypotenuse! To plot z = -3 + 4i: (1) Identify real part a = -3 and imaginary part b = 4 (the coefficient of i). (2) Plot at point (-3, 4): start at origin, move 3 units left (negative real), then 4 units up (positive imaginary). (3) Calculate modulus: |z| = square root of ((-3) squared + 4 squared) = square root of (9 + 16) = square root of 25 = 5. The point is 5 units from the origin! Choice C correctly identifies the point as (-3, 4) and calculates |z| = 5 using the distance formula. Choice B incorrectly swaps the coordinates to (4, -3) instead of (-3, 4)—remember, real part goes horizontal, imaginary vertical! Choice B also miscalculates |z| = 7 by adding |-3| + |4| instead of using the Pythagorean theorem. Choice D has the wrong sign on the real part. Plotting complex numbers recipe: (1) From a + bi, a is horizontal coordinate, b is vertical. (2) Plot at (a, b). (3) Modulus = square root of (a squared + b squared). Always check your signs carefully—negative real means left, negative imaginary means down!

Question 2

Complete the square to write the circle equation x2+y2−8x+6y+9=0x^2+y^2-8x+6y+9=0x2+y2−8x+6y+9=0 in standard form. Which standard-form equation is correct?

  1. (x−4)2+(y+3)2=16(x-4)^2+(y+3)^2=16(x−4)2+(y+3)2=16 (correct answer)
  2. (x+4)2+(y−3)2=16(x+4)^2+(y-3)^2=16(x+4)2+(y−3)2=16
  3. (x−4)2+(y+3)2=4(x-4)^2+(y+3)^2=4(x−4)2+(y+3)2=4
  4. (x−8)2+(y+6)2=9(x-8)^2+(y+6)^2=9(x−8)2+(y+6)2=9

Explanation: This question tests your understanding of how to derive circle equations using the Pythagorean Theorem (distance formula) or find a circle's center and radius by completing the square. When a circle equation is in general form x² + y² + Dx + Ey + F = 0, we complete the square in both x and y to reveal the center and radius: group x-terms (x² + Dx) and complete the square by adding (D/2)², do the same for y-terms with (E/2)², then rearrange to get (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F. The center is (-D/2, -E/2) and radius is √[(D/2)² + (E/2)² - F]. It's completing the square twice, once for each variable! Starting with x² + y² - 8x + 6y + 9 = 0, move 9 to get x² - 8x + y² + 6y = -9, complete square for x: (x - 4)² - 16, for y: (y + 3)² - 9, so (x - 4)² + (y + 3)² = -9 + 16 + 9 = 16. Choice A correctly completes the square to identify the standard form (x - 4)² + (y + 3)² = 16. A distractor like choice B might forget the signs when completing the square, using +4 for x instead of -4, but gently remember D = -8 so -D/2 = 4 for the center's x-coordinate. Completing the square for circles: (1) Group x-terms together and y-terms together: (x² + Dx) + (y² + Ey) = -F, (2) Complete square in x by adding (D/2)² to both sides, (3) Complete square in y by adding (E/2)² to both sides, (4) Factor: (x + D/2)² + (y + E/2)² = (D/2)² + (E/2)² - F, (5) Read center as (-D/2, -E/2) and radius as √[right side].

Question 3

In the expansion of (x+y)6(x+y)^6(x+y)6, what is the coefficient of x3y3x^3 y^3x3y3? (Pascal's Triangle row 6: 1,6,15,20,15,6,11,6,15,20,15,6,11,6,15,20,15,6,1.)

  1. 30
  2. 15
  3. 20 (correct answer)
  4. 18

Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x+y)n(x + y)^n(x+y)n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (x+y)n(x + y)^n(x+y)n expands to a sum of n+1n+1n+1 terms where coefficients come from row nnn of Pascal's Triangle, xxx-powers decrease from nnn to 0, and yyy-powers increase from 0 to nnn. For example, (x+y)4(x + y)^4(x+y)4 uses row 4 of Pascal's Triangle (1, 4, 6, 4, 1) to give: 1⋅x4+4⋅x3y+6⋅x2y2+4⋅xy3+1⋅y41 \cdot x^4 + 4 \cdot x^3 y + 6 \cdot x^2 y^2 + 4 \cdot x y^3 + 1 \cdot y^41⋅x4+4⋅x3y+6⋅x2y2+4⋅xy3+1⋅y4. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! In (x+y)6(x+y)^6(x+y)6 with row 6 (1,6,15,20,15,6,1), the term x3y3x^3 y^3x3y3 uses the fourth coefficient, 20, as choice C states. Choice A (15) is for adjacent terms like x4y2x^4 y^2x4y2 or x2y4x^2 y^4x2y4, but for equal exponents in even nnn, it's the middle peak. The expansion recipe using Pascal's Triangle: (1) Identify nnn (the exponent on the binomial), (2) Write or construct row nnn of Pascal's Triangle—you'll have n+1n+1n+1 numbers, (3) Create n+1n+1n+1 terms: first has coefficient 1 and is xnx^nxn, last has coefficient 1 and is yny^nyn, middle terms use Pascal's row with decreasing xxx-powers and increasing yyy-powers, (4) Write it out: [1st coefficient]·xnx^nxn + [2nd coefficient]·xn−1yx^{n-1} yxn−1y + [3rd coefficient]·xn−2y2x^{n-2} y^2xn−2y2 + ... The pattern is systematic and reliable!

Question 4

A tank is being filled. The volume of water VVV (in liters) is recorded every minute ttt.

ttt (min)01234
VVV (L)1013172228

Calculate the rate of change over each 1-minute interval. Is the rate of change constant?

  1. No; ΔV\Delta VΔV values (3, 4, 5, 6) are not all equal, so the rate is non-constant (nonlinear). (correct answer)
  2. Yes; the increases are 3, 4, 5, 6 so the constant rate is 4.5 L/min.
  3. Yes; because time increases by 1 each row, the rate must be constant.
  4. No; but only the first interval matters, and it is 3 L/min so the rate is constant at 3 L/min.

Explanation: This question tests your ability to recognize when a relationship has constant rate of change—the defining characteristic of linear functions. Constant rate of change means that for every unit increase in x, y changes by the same amount every time: if Δy/Δx = 5 for one interval and also 5 for every other equal interval, the rate is constant at 5. This constant rate is exactly what makes a function linear (y = mx + b where m is that constant rate). Only linear functions have this property—quadratics, exponentials, and other nonlinear functions have rates that vary at different x-values! Here, ΔV for Δt=1 are 3,4,5,6, which vary, so non-constant rate. Choice B correctly identifies the non-constant rate because the ΔV values are not equal, indicating nonlinear. A mistake like in Choice A is averaging varying differences, but for constant rate, they must be identical, not averaged! The three-method constant rate test: METHOD 1 (from table): Calculate Δy/Δx for each pair of consecutive points with equal Δx. All equal? Constant rate. Vary? Non-constant. METHOD 2 (from graph): Is it a straight line? Yes = constant rate. Curved? Non-constant. METHOD 3 (from formula): Is it y = mx + b form? Yes = constant rate m. Any other form (x², b^x, etc.)? Non-constant. Pick the method matching your representation! Don't confuse constant RATE with constant RATIO: constant rate (Δy/Δx equal) characterizes linear functions, constant ratio (y₂/y₁ equal) characterizes exponential functions. Check BOTH in a table: if differences are 3, 3, 3 → linear with rate 3. If ratios are 2, 2, 2 → exponential with base 2. If neither constant → some other type. Knowing which pattern to look for prevents confusing linear with exponential growth!

Question 5

Use log properties to evaluate without a calculator: log⁡3(819).\log_3\left(\frac{81}{9}\right).log3​(981​). You may use log⁡b(xy)=log⁡b(x)−log⁡b(y)\log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y)logb​(yx​)=logb​(x)−logb​(y) and log⁡b(bk)=k\log_b(b^k)=klogb​(bk)=k.

  1. 000
  2. 111
  3. 222 (correct answer)
  4. 444

Explanation: This question tests your understanding of the three fundamental logarithm properties—product, quotient, and power rules—that let you expand complex logarithmic expressions into sums and differences or condense multiple logs into a single logarithm. The quotient property states log⁡b(xy)=log⁡b(x)−log⁡b(y)\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y)logb​(yx​)=logb​(x)−logb​(y), and the key evaluation fact is log⁡b(bk)=k\log_b(b^k) = klogb​(bk)=k. To evaluate log⁡3(819)\log_3\left(\frac{81}{9}\right)log3​(981​): (1) Apply quotient property: log⁡3(81)−log⁡3(9)\log_3(81) - \log_3(9)log3​(81)−log3​(9), (2) Recognize 81=3481 = 3^481=34 and 9=329 = 3^29=32 as powers of base 3, (3) Use log⁡b(bk)=k\log_b(b^k) = klogb​(bk)=k: log⁡3(34)−log⁡3(32)=4−2=2\log_3(3^4) - \log_3(3^2) = 4 - 2 = 2log3​(34)−log3​(32)=4−2=2. Choice C correctly evaluates to 2 by recognizing both 81 and 9 as powers of 3 and applying the quotient property. Choice A would be the result of incorrectly thinking 81/9=181/9 = 181/9=1, but 81/9=981/9 = 981/9=9, not 1! Evaluation strategy for quotients: (1) Apply quotient property to separate into difference of logs, (2) Express each number as power of the base if possible: for base 3, know that 31=33^1=331=3, 32=93^2=932=9, 33=273^3=2733=27, 34=813^4=8134=81, (3) Use log⁡b(bk)=k\log_b(b^k) = klogb​(bk)=k to evaluate each term, (4) Subtract to get final answer. Alternative shortcut: simplify the fraction first (819=9=32\frac{81}{9} = 9 = 3^2981​=9=32), then evaluate log⁡3(32)=2\log_3(3^2) = 2log3​(32)=2 directly!

Question 6

Two sequences are defined as follows: Sequence A: u1=4u_1 = 4u1​=4, un=un−1+3u_n = u_{n-1} + 3un​=un−1​+3 for n≥2n \geq 2n≥2. Sequence B: v1=4v_1 = 4v1​=4, vn=3vn−1v_n = 3v_{n-1}vn​=3vn−1​ for n≥2n \geq 2n≥2. When comparing these sequences as functions, which statement is true?

  1. Both sequences can be written as explicit linear functions of nnn
  2. Sequence A grows exponentially while Sequence B grows linearly
  3. Sequence A can be written as u(n)=3n+1u(n) = 3n + 1u(n)=3n+1 and Sequence B grows exponentially (correct answer)
  4. Both sequences have the same rate of growth for large values of nnn

Explanation: Sequence A is arithmetic: u(n) = 4 + 3(n-1) = 3n + 1, which is linear. Sequence B is geometric: v(n) = 4·3^(n-1), which is exponential. Choice A is wrong because B is exponential, not linear. Choice B reverses the growth types. Choice D is wrong because exponential growth eventually dominates linear growth.

Question 7

For the linear function f(x)=3x+2f(x)=3x+2f(x)=3x+2, verify over equal unit intervals (h=1h=1h=1) that the differences are constant by computing f(0),f(1),f(2),f(3)f(0), f(1), f(2), f(3)f(0),f(1),f(2),f(3) and the consecutive differences. Which option correctly demonstrates the constant-difference property and identifies the associated sequence type at integer inputs?

  1. f(0)=2,f(1)=5,f(2)=8,f(3)=11f(0)=2, f(1)=5, f(2)=8, f(3)=11f(0)=2,f(1)=5,f(2)=8,f(3)=11; differences: 3,3,33,3,33,3,3 (constant). So {f(n)}\{f(n)\}{f(n)} is arithmetic. (correct answer)
  2. f(0)=2,f(1)=6,f(2)=18,f(3)=54f(0)=2, f(1)=6, f(2)=18, f(3)=54f(0)=2,f(1)=6,f(2)=18,f(3)=54; ratios: 3,3,33,3,33,3,3 (constant). So {f(n)}\{f(n)\}{f(n)} is geometric.
  3. f(0)=2,f(1)=5,f(2)=9,f(3)=14f(0)=2, f(1)=5, f(2)=9, f(3)=14f(0)=2,f(1)=5,f(2)=9,f(3)=14; differences: 3,4,53,4,53,4,5 (constant). So {f(n)}\{f(n)\}{f(n)} is arithmetic.
  4. f(0)=2,f(1)=5,f(2)=8,f(3)=11f(0)=2, f(1)=5, f(2)=8, f(3)=11f(0)=2,f(1)=5,f(2)=8,f(3)=11; ratios: 52,85,118\tfrac{5}{2},\tfrac{8}{5},\tfrac{11}{8}25​,58​,811​ (constant). So {f(n)}\{f(n)\}{f(n)} is geometric.

Explanation: This question tests your understanding of a fundamental distinction: linear functions grow by adding the same amount over equal intervals (constant differences), while exponential functions grow by multiplying by the same factor over equal intervals (constant ratios). For any linear function f(x) = mx + b, the change over an interval of length h is constant: f(x + h) - f(x) = [m(x + h) + b] - [mx + b] = mh + b - b = mh, which depends only on the interval length h and slope m, not on where you start (x). This constant difference mh means moving h units right always adds the same amount to the function value. For h = 1, you always add m (the slope). This additive pattern defines linearity! Let's compute f(x) = 3x + 2 at x = 0, 1, 2, 3: f(0) = 3(0) + 2 = 2, f(1) = 3(1) + 2 = 5, f(2) = 3(2) + 2 = 8, f(3) = 3(3) + 2 = 11. The differences are: 5 - 2 = 3, 8 - 5 = 3, 11 - 8 = 3. All differences equal 3, which is the slope m! Choice A correctly computes these values and differences, showing they're constant at 3, and correctly identifies that {f(n)} forms an arithmetic sequence. Choice B shows values for an exponential function (2, 6, 18, 54) with constant ratios of 3, not the linear function f(x) = 3x + 2. Choice C has an arithmetic error: f(2) = 3(2) + 2 = 8, not 9, leading to incorrect differences of 3, 4, 5. Choice D correctly computes the values but then incorrectly calculates ratios instead of differences - for a linear function, ratios are not constant unless the function passes through the origin. To determine if a function is linear or exponential from a table: (1) check if x-values have equal spacing (like 0, 1, 2, 3 or 0, 5, 10, 15), (2) calculate differences between consecutive y-values: y₂ - y₁, y₃ - y₂, y₄ - y₃, (3) calculate ratios: y₂/y₁, y₃/y₂, y₄/y₃, (4) if differences are constant → linear (slope = that constant difference per unit interval), if ratios are constant → exponential (base = that constant ratio per unit interval). Can't be both unless the function is constant! Why this matters: the growth pattern reveals the function type and lets you predict future values. If differences are constant at 5, the next value is 'current + 5.' If ratios are constant at 1.2, the next value is 'current × 1.2.' Linear growth is steady and predictable (add same amount), exponential growth accelerates (each addition is larger because it's a percentage of a growing base). Understanding these patterns is key to recognizing linear vs exponential in data, formulas, and real-world contexts!

Question 8

Verify the inverse relationship between the temperature conversion formulas C(F)=59(F−32)C(F)=\dfrac{5}{9}(F-32)C(F)=95​(F−32) and F(C)=95C+32F(C)=\dfrac{9}{5}C+32F(C)=59​C+32 by showing that C(F(C))=CC(F(C))=CC(F(C))=C and F(C(F))=FF(C(F))=FF(C(F))=F.​

  1. C(F(C))=59(95C+32)=C+1609≠CC(F(C))=\dfrac{5}{9}\left(\dfrac{9}{5}C+32\right)=C+\dfrac{160}{9}\ne CC(F(C))=95​(59​C+32)=C+9160​=C, so they are not inverses.
  2. C(F(C))=CC(F(C))=CC(F(C))=C; therefore the formulas are inverses without needing to check F(C(F))F(C(F))F(C(F)).
  3. C(F(C))=59(95C)+32=C+32≠CC(F(C))=\dfrac{5}{9}\left(\dfrac{9}{5}C\right)+32=C+32\ne CC(F(C))=95​(59​C)+32=C+32=C and F(C(F))=FF(C(F))=FF(C(F))=F, so they are not inverses.
  4. C(F(C))=59((95C+32)−32)=59⋅95C=CC(F(C))=\dfrac{5}{9}\left(\left(\dfrac{9}{5}C+32\right)-32\right)=\dfrac{5}{9}\cdot\dfrac{9}{5}C=CC(F(C))=95​((59​C+32)−32)=95​⋅59​C=C and F(C(F))=95(59(F−32))+32=F−32+32=FF(C(F))=\dfrac{9}{5}\left(\dfrac{5}{9}(F-32)\right)+32=F-32+32=FF(C(F))=59​(95​(F−32))+32=F−32+32=F, so they are inverses. (correct answer)

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! For C(F(C)) = C((9/5)C + 32) = (5/9)((9/5)C + 32 - 32) = (5/9)(9/5)C = C ✓, and F(C(F)) = F((5/9)(F-32)) = (9/5)((5/9)(F-32)) + 32 = F - 32 + 32 = F ✓. Both compositions return the original input! Choice A correctly verifies both compositions (C(F(C)) = C and F(C(F)) = F) and confirms the inverse relationship. Choice B forgets to subtract 32 before multiplying by 5/9, Choice C only checks one direction which is insufficient, and Choice D adds 32 instead of subtracting it in C(F(C)). The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable!

Question 9

A community garden is deciding how many tomato plants and pepper plants to grow. Let xxx = number of tomato plants and yyy = number of pepper plants.

Constraints:

  • Each tomato plant needs 3 square feet; each pepper plant needs 2 square feet; at most 120 square feet are available: 3x+2y≤1203x+2y\le 1203x+2y≤120.
  • They want at least 40 plants total: x+y≥40x+y\ge 40x+y≥40.
  • They must plant at least 10 pepper plants: y≥10y\ge 10y≥10.
  • Plants must be whole numbers and cannot be negative.

Which option is a viable plan?

  1. (x,y)=(25,15)(x,y)=(25,15)(x,y)=(25,15) (correct answer)
  2. (x,y)=(35,5)(x,y)=(35,5)(x,y)=(35,5)
  3. (x,y)=(30,12.5)(x,y)=(30,12.5)(x,y)=(30,12.5)
  4. (x,y)=(50,10)(x,y)=(50,10)(x,y)=(50,10)

Explanation: This question tests your ability to translate real-world limitations into mathematical constraints (equations and inequalities) and determine whether potential solutions are viable—meaning they satisfy all constraints and make sense in context. A constraint is a limitation or requirement: 'budget at most 500′becomescost≤500,′needatleast10units′becomesquantity≥10.ThesystemofALLconstraintsdefinesthefeasibleregion—thesetofallsolutionsthatwork.Asolutionisviableifitliesinthisfeasibleregion(satisfieseverysingleinequality/equation)ANDmakesreal−worldsense(nonegativequantities,wholeunitswhenneeded,etc.).Evenoneviolationmakesitnonviable!Let′scheckeachoption:ForA(25,15):3(25)+2(15)=75+30=105≤120✓,25+15=40≥40✓,15≥10✓,wholenumbers✓.ForB(35,5):3(35)+2(5)=105+10=115≤120✓,35+5=40≥40✓,5<10✗.ForC(30,12.5):12.5isnotawholenumber✗.ForD(50,10):3(50)+2(10)=150+20=170>120✗.ChoiceAcorrectlyidentifies(25,15)astheonlyviableplanthatsatisfiesallconstraintsincludingthewholenumberrequirement.OptionsB,C,andDeachviolateatleastoneconstraint:Bdoesn′tmeetminimumpeppers,Cusesfractionalplants,andDexceedsspacelimit.Constraintidentificationfromcontext:(1)listeverylimitationmentioned(′budget500' becomes cost ≤ 500, 'need at least 10 units' becomes quantity ≥ 10. The system of ALL constraints defines the feasible region—the set of all solutions that work. A solution is viable if it lies in this feasible region (satisfies every single inequality/equation) AND makes real-world sense (no negative quantities, whole units when needed, etc.). Even one violation makes it nonviable! Let's check each option: For A (25,15): 3(25) + 2(15) = 75 + 30 = 105 ≤ 120 ✓, 25 + 15 = 40 ≥ 40 ✓, 15 ≥ 10 ✓, whole numbers ✓. For B (35,5): 3(35) + 2(5) = 105 + 10 = 115 ≤ 120 ✓, 35 + 5 = 40 ≥ 40 ✓, 5 < 10 ✗. For C (30,12.5): 12.5 is not a whole number ✗. For D (50,10): 3(50) + 2(10) = 150 + 20 = 170 > 120 ✗. Choice A correctly identifies (25,15) as the only viable plan that satisfies all constraints including the whole number requirement. Options B, C, and D each violate at least one constraint: B doesn't meet minimum peppers, C uses fractional plants, and D exceeds space limit. Constraint identification from context: (1) list every limitation mentioned ('budget 500′becomescost≤500,′needatleast10units′becomesquantity≥10.ThesystemofALLconstraintsdefinesthefeasibleregion—thesetofallsolutionsthatwork.Asolutionisviableifitliesinthisfeasibleregion(satisfieseverysingleinequality/equation)ANDmakesreal−worldsense(nonegativequantities,wholeunitswhenneeded,etc.).Evenoneviolationmakesitnonviable!Let′scheckeachoption:ForA(25,15):3(25)+2(15)=75+30=105≤120✓,25+15=40≥40✓,15≥10✓,wholenumbers✓.ForB(35,5):3(35)+2(5)=105+10=115≤120✓,35+5=40≥40✓,5<10✗.ForC(30,12.5):12.5isnotawholenumber✗.ForD(50,10):3(50)+2(10)=150+20=170>120✗.ChoiceAcorrectlyidentifies(25,15)astheonlyviableplanthatsatisfiesallconstraintsincludingthewholenumberrequirement.OptionsB,C,andDeachviolateatleastoneconstraint:Bdoesn′tmeetminimumpeppers,Cusesfractionalplants,andDexceedsspacelimit.Constraintidentificationfromcontext:(1)listeverylimitationmentioned(′budgetX,' 'time ≤ Y hours,' 'need ≥ Z units'), (2) translate using key phrases: 'at most' → ≤, 'at least' → ≥, 'exactly' → =, 'more than' → >, 'less than' → <, (3) don't forget implicit constraints like x ≥ 0, y ≥ 0 (can't be negative) or x, y integers (if discrete), (4) write the complete system. Missing even one constraint can make you accept infeasible solutions!

Question 10

Use factoring to determine where the function equals zero: g(x)=x2−25g(x)=x^2-25g(x)=x2−25. What are the zeros, and what x-intercepts does this give for the graph?

  1. g(x)=(x−25)(x+1)g(x)=(x-25)(x+1)g(x)=(x−25)(x+1); zeros: x=25,−1x=25,-1x=25,−1; x-intercepts: (25,0),(−1,0)(25,0),(-1,0)(25,0),(−1,0)
  2. g(x)=(x−5)2g(x)=(x-5)^2g(x)=(x−5)2; zero: x=5x=5x=5; x-intercept: (5,0)(5,0)(5,0)
  3. g(x)=(x+5)(x+5)g(x)=(x+5)(x+5)g(x)=(x+5)(x+5); zero: x=−5x=-5x=−5; x-intercept: (−5,0)(-5,0)(−5,0)
  4. g(x)=(x−5)(x+5)g(x)=(x-5)(x+5)g(x)=(x−5)(x+5); zeros: x=5,−5x=5,-5x=5,−5; x-intercepts: (5,0),(−5,0)(5,0),(-5,0)(5,0),(−5,0) (correct answer)

Explanation: This question tests your ability to factor quadratic expressions, specifically difference of squares, and use the factored form to identify zeros (x-intercepts) of the function—essential for graphing and solving quadratic equations. Factoring a quadratic into form g(x) = (x - r)(x - s) reveals the zeros immediately: set the factored expression equal to zero and use the zero product property (if a product equals zero, at least one factor must equal zero). Setting (x - r) = 0 gives x = r, and setting (x - s) = 0 gives x = s, so the zeros are r and s. These are the x-intercepts of the parabola—points (r, 0) and (s, 0) where the graph crosses the x-axis. Factoring transforms the quadratic from a form where zeros are hidden (standard form) to a form where they're obvious (factored form)! To factor x² - 25 and find zeros: recognize it as a difference of squares, a² - b² = (a - b)(a + b), where a = x, b = 5. So (x - 5)(x + 5). Find zeros: x - 5 = 0 gives x = 5, x + 5 = 0 gives x = -5. X-intercepts: (5, 0), (-5, 0). Great job spotting the pattern! Choice B correctly factors using difference of squares and identifies the zeros accurately. Choice C mistakes it for a perfect square trinomial, but x² - 25 isn't (x - 5)² = x² - 10x + 25; there's no middle term, so it's difference of squares, not a repeated factor. Always check if it's a² - b²! Factoring difference of squares is a quick shortcut: x² - c = (x - √c)(x + √c) if c > 0. Example: x² - 16 = (x - 4)(x + 4), zeros ±4. For sums like x² + 25, it doesn't factor over reals. Remember the pattern to save time!

Question 11

The expression 9x2+369x^2 + 369x2+36 can be factored over the complex numbers as 9(x+ai)(x−ai)9(x + ai)(x - ai)9(x+ai)(x−ai) where aaa is a positive real number. What is the value of aaa?

  1. 222 (correct answer)
  2. 666
  3. 333
  4. 444

Explanation: First, factor out the common factor: 9x2+36=9(x2+4)9x^2 + 36 = 9(x^2 + 4)9x2+36=9(x2+4). To factor x2+4x^2 + 4x2+4 over the complex numbers, we need x2+4=x2−(−4)=x2−(2i)2=(x+2i)(x−2i)x^2 + 4 = x^2 - (-4) = x^2 - (2i)^2 = (x + 2i)(x - 2i)x2+4=x2−(−4)=x2−(2i)2=(x+2i)(x−2i). Therefore, 9x2+36=9(x+2i)(x−2i)9x^2 + 36 = 9(x + 2i)(x - 2i)9x2+36=9(x+2i)(x−2i), which matches the form 9(x+ai)(x−ai)9(x + ai)(x - ai)9(x+ai)(x−ai) with a=2a = 2a=2. Choice B results from confusing the coefficient 36 with the value under the square root. Choice C comes from taking the square root of 9. Choice D results from taking the square root of 16 instead of 4.

Question 12

Find the expansion of (2x−3)3(2x-3)^3(2x−3)3 using the Binomial Theorem.

  1. 8x3−36x2+54x−278x^3-36x^2+54x-278x3−36x2+54x−27 (correct answer)
  2. 8x3−18x2+54x−278x^3-18x^2+54x-278x3−18x2+54x−27
  3. 8x3+36x2−54x+278x^3+36x^2-54x+278x3+36x2−54x+27
  4. 4x3−36x2+54x−274x^3-36x^2+54x-274x3−36x2+54x−27

Explanation: This question tests your understanding of the Binomial Theorem—a formula for expanding (x + y)^n using coefficients from Pascal's Triangle, which saves enormous time compared to multiplying out by hand! The Binomial Theorem says (x + y)^n expands to a sum of n+1 terms where coefficients come from row n of Pascal's Triangle, x-powers decrease from n to 0, and y-powers increase from 0 to n. For example, (x + y)⁴ uses row 4 of Pascal's Triangle (1, 4, 6, 4, 1) to give: 1·x⁴ + 4·x³y + 6·x²y² + 4·xy³ + 1·y⁴. Each term has exponents summing to 4, and coefficients from the triangle make this work perfectly! For (2x-3)³, we use row 3 of Pascal's Triangle (1, 3, 3, 1) and treat it as (2x + (-3))³. The expansion gives: 1·(2x)³·(-3)⁰ + 3·(2x)²·(-3)¹ + 3·(2x)¹·(-3)² + 1·(2x)⁰·(-3)³ = 8x³ + 3·4x²·(-3) + 3·2x·9 + 1·(-27) = 8x³ - 36x² + 54x - 27. Choice A correctly expands with proper coefficients: 8x³ - 36x² + 54x - 27. Choice B has -18x² instead of -36x² (error in calculating 3·4·(-3)), Choice C has wrong signs throughout, and Choice D starts with 4x³ instead of 8x³ (error in calculating (2x)³). The expansion recipe using Pascal's Triangle: (1) Identify n (the exponent on the binomial), (2) Write or construct row n of Pascal's Triangle—you'll have n+1 numbers, (3) Create n+1 terms: first has coefficient 1 and is x^n, last has coefficient 1 and is y^n, middle terms use Pascal's row with decreasing x-powers and increasing y-powers, (4) Write it out: [1st coefficient]·x^n + [2nd coefficient]·x^(n-1)·y + [3rd coefficient]·x^(n-2)·y² + ... The pattern is systematic and reliable!

Question 13

In geometry, the area of a circle is A=πr2A=\pi r^2A=πr2, where rrr is the radius. Rearrange A=πr2A=\pi r^2A=πr2 to make rrr the subject (use inverse operations as with solving an equation).

  1. r=πAr=\sqrt{\pi A}r=πA​
  2. r=Aπr=\dfrac{A}{\pi}r=πA​
  3. r=Aπr=\sqrt{\dfrac{A}{\pi}}r=πA​​ (correct answer)
  4. r=Aπ2r=\dfrac{A}{\pi^2}r=π2A​

Explanation: This question tests your ability to rearrange formulas to solve for a specific variable—essential for using formulas flexibly in science, engineering, and real-world problem-solving. More complex rearrangements may require advanced techniques: if your target variable is squared (like r² in A = πr²), you'll need square roots (r = √(A/π)). If it appears in a denominator (like f in 1/f = 1/a + 1/b), you'll need to clear fractions first. If it appears with different powers (like t in s = ut + (1/2)at²), you may need the quadratic formula! The complexity of the rearrangement depends on how the variable appears in the formula. Starting with A = πr², we need to isolate r. First divide both sides by π: A/π = r². Then take the square root of both sides: √(A/π) = r, which gives us r = √(A/π). Choice A correctly isolates r through division by π and taking the square root to get r = √(A/π). Choice B incorrectly divides by π² instead of π, Choice C forgets to take the square root (leaving r²), and Choice D incorrectly multiplies inside the square root instead of dividing. Watch for variables appearing multiple times: if your target variable appears in multiple places (like x in xy + xz = w), factor it out first: x(y + z) = w, then x = w/(y + z). If you don't factor, you'll struggle to isolate! Also, when taking square roots of a variable, remember ± if the formula context allows both positive and negative (though often context restricts to positive only, like radius r ≥ 0). Physics and geometry formulas usually want positive values only!

Question 14

Two functions f(x)=x2−4x+5f(x) = x^2 - 4x + 5f(x)=x2−4x+5 and g(x)=−x2+6x−1g(x) = -x^2 + 6x - 1g(x)=−x2+6x−1 are graphed together. At their intersection points, what is the relationship between the solutions of f(x)=g(x)f(x) = g(x)f(x)=g(x) and the solutions of 2x2−10x+6=02x^2 - 10x + 6 = 02x2−10x+6=0?

  1. They are reciprocals of each other due to the algebraic manipulation involved
  2. They are different solutions since the second equation has different coefficients
  3. They are related by a factor of 2, so one set of solutions is twice the other
  4. They are the same solutions since both equations represent the intersection condition (correct answer)

Explanation: When you encounter questions about function intersections, remember that intersection points occur where two functions have equal output values for the same input values. This means you're looking for x-values where f(x)=g(x)f(x) = g(x)f(x)=g(x). To find where these functions intersect, you set them equal: x2−4x+5=−x2+6x−1x^2 - 4x + 5 = -x^2 + 6x - 1x2−4x+5=−x2+6x−1. Moving all terms to one side gives you x2−4x+5+x2−6x+1=0x^2 - 4x + 5 + x^2 - 6x + 1 = 0x2−4x+5+x2−6x+1=0, which simplifies to 2x2−10x+6=02x^2 - 10x + 6 = 02x2−10x+6=0. This is exactly the second equation given in the question! The solutions to both equations are identical because they represent the same mathematical condition: finding the x-coordinates where the graphs intersect. Choice A is incorrect because reciprocals aren't involved in this algebraic process—you're simply rearranging terms, not inverting values. Choice B misses the fundamental concept: having different coefficients doesn't mean different solutions when the equations are mathematically equivalent through algebraic manipulation. Choice C incorrectly assumes a proportional relationship between the solution sets, but there's no factor of 2 connecting them—the solutions are exactly the same numbers. The correct answer is D because both equations represent identical intersection conditions, just written in different forms. Study tip: When comparing equations that might represent the same relationships, always check if one can be algebraically transformed into the other. If so, their solutions will be identical, regardless of how different the coefficients initially appear.

Question 15

A phone battery retains a constant percent of its charge every hour. The charge levels are:

Hour ttt: 0, 1, 2, 3 Charge (%): 100, 92, 84.64, 77.8688

What percent does the battery decrease per hour?

  1. Neither; it is linear because it drops by 8% each hour.
  2. 92% decay per hour.
  3. 8% growth per hour.
  4. 8% decay per hour. (correct answer)

Explanation: This question tests your ability to recognize exponential growth or decay—situations where a quantity changes by a constant percent (not constant amount) per time interval. Constant percent growth means multiplying by the same factor greater than 1 each interval: if something grows 5% per year, each year's value is 105% of the previous (multiply by 1.05). Constant percent decay means multiplying by a factor between 0 and 1: if something decreases 15% per year, each year retains 85% of previous (multiply by 0.85). The key test: calculate ratios of consecutive values. If ratios are constant, it's exponential with that ratio as the growth/decay factor! To check, compute the ratios from the table: 92/100 = 0.92, 84.64/92 = 0.92, and 77.8688/84.64 = 0.92, showing a constant ratio of 0.92, fitting exponential decay. Choice A correctly identifies 8% decay per hour through constant ratios of 0.92 (since 1 - 0.92 = 0.08 or 8%). One distractor suggests 92% decay, but that's confusing the factor with the rate—the rate is the change, so 8%; you're doing great learning this distinction! The ratio test for exponential: (1) from table, divide consecutive y-values: y₂/y₁, y₃/y₂, y₄/y₃, (2) if all equal → exponential with that ratio as base b, (3) if b > 1 → growth; if 0 < b < 1 → decay, (4) calculate percent rate: r = b - 1, convert to percent. Example: ratios all 1.06 → base 1.06 → growth → rate = 1.06 - 1 = 0.06 = 6% per interval. This systematic check identifies exponential patterns reliably! Don't confuse exponential with linear: linear adds the same amount each time (constant differences like +50, +50, +50), exponential multiplies by same factor (constant ratios like ×1.1, ×1.1, ×1.1). Both are patterns of regular change, but different mechanisms! Check both: if differences constant → linear. If ratios constant → exponential. Usually only one pattern holds. The language helps too: 'grows by $50 per year' = linear (additive), 'grows by 5% per year' = exponential (multiplicative)!

Question 16

Use polynomial long division to rewrite 4x3−x2+2x−1x+1\frac{4x^3-x^2+2x-1}{x+1}x+14x3−x2+2x−1​ in the form q(x)+r(x)x+1q(x)+\frac{r(x)}{x+1}q(x)+x+1r(x)​, where deg⁡(r)<1\deg(r)<1deg(r)<1.

  1. 4x2−5x+7+−8x+14x^2-5x+7+\dfrac{-8}{x+1}4x2−5x+7+x+1−8​ (correct answer)
  2. 4x2−5x+7+8x+14x^2-5x+7+\dfrac{8}{x+1}4x2−5x+7+x+18​
  3. 4x2+5x+7+−8x+14x^2+5x+7+\dfrac{-8}{x+1}4x2+5x+7+x+1−8​
  4. 4x2−5x+−8xx+14x^2-5x+\dfrac{-8x}{x+1}4x2−5x+x+1−8x​

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. Let's divide (4x³ - x² + 2x - 1) by (x + 1): First, 4x³ ÷ x = 4x², multiply back: 4x²(x + 1) = 4x³ + 4x², subtract to get -5x² + 2x - 1. Next, -5x² ÷ x = -5x, multiply back: -5x(x + 1) = -5x² - 5x, subtract to get 7x - 1. Finally, 7x ÷ x = 7, multiply back: 7(x + 1) = 7x + 7, subtract to get -8. Choice A correctly shows quotient 4x² - 5x + 7 and remainder -8, giving us 4x² - 5x + 7 + (-8)/(x + 1). Choice B has remainder 8 instead of -8, missing the negative sign from the final subtraction: -1 - 7 = -8. Before dividing, always check if you can factor and simplify—here we can't, so long division is necessary. The degree requirement tells you to stop when remainder degree < divisor degree: -8 has degree 0, which is less than 1, so we're done!

Question 17

Pipe A can drain a tank in 5 hours. Pipe B can drain the same tank in 10 hours. If both pipes are opened at the same time, how long will it take to drain the tank?

Write and solve a rational equation.

Let ttt = time in hours to drain the tank together.

  1. Set up 15−110=1t\frac{1}{5}-\frac{1}{10}=\frac{1}{t}51​−101​=t1​. Then t=10t=10t=10 hours, so it takes 10 hours.
  2. Set up 1t=510\frac{1}{t}=\frac{5}{10}t1​=105​. Then t=2t=2t=2 hours, so it takes 2 hours.
  3. Set up 15+110=1t\frac{1}{5}+\frac{1}{10}=\frac{1}{t}51​+101​=t1​. Then t=103t=\frac{10}{3}t=310​ hours, so it takes about 3.33 hours. (correct answer)
  4. Set up 5+102=t\frac{5+10}{2}=t25+10​=t. Then t=7.5t=7.5t=7.5 hours, so it takes 7.5 hours.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Work-rate problems lead to rational equations: if one worker completes a job in time t₁ and another in t₂, their combined rate is 1/t₁ + 1/t₂ (adding rates), which equals 1/t_combined. The reciprocals represent 'fraction of job per hour,' and adding these fractions gives the combined rate. Solving these rational equations requires finding LCD and often produces fractional time answers that make sense: 3.4 hours = 3 hours 24 minutes. For draining the tank, set up 1/5 + 1/10 = 1/t; common denominator 10 gives 2/10 + 1/10 = 3/10 = 1/t, so t = 10/3 ≈ 3.33 hours, meaning both pipes drain the tank in about 3 hours 20 minutes. Choice A correctly sets up the equation by adding the draining rates and solves to find t = 10/3 hours. A distractor like choice B subtracts rates, which would apply if one was filling instead of both draining. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 18

Does the table show a constant rate of change between xxx and yyy? Explain using Δy/Δx\Delta y/\Delta xΔy/Δx.

Table (equal xxx-intervals of 1): xxx: 0, 1, 2, 3 yyy: 4, 7, 10, 13

  1. No; the differences in yyy are 3, 4, 5 so the rate changes.
  2. Yes; Δy/Δx=3\Delta y/\Delta x = 3Δy/Δx=3 for each interval, so the rate is constant and the relationship is linear. (correct answer)
  3. Yes; because y/xy/xy/x is constant for all xxx values.
  4. No; Δy/Δx=4\Delta y/\Delta x = 4Δy/Δx=4 for each interval, so it is not linear.

Explanation: This question tests your ability to recognize when a relationship has constant rate of change—the defining characteristic of linear functions. Constant rate of change means that for every unit increase in x, y changes by the same amount every time: if Δy/Δx = 5 for one interval and also 5 for every other equal interval, the rate is constant at 5. This constant rate is exactly what makes a function linear (y = mx + b where m is that constant rate). Only linear functions have this property—quadratics, exponentials, and other nonlinear functions have rates that vary at different x-values! For this table, calculate the rates: from x=0 to 1, Δy/Δx = (7-4)/1 = 3; from x=1 to 2, (10-7)/1 = 3; from x=2 to 3, (13-10)/1 = 3, so the rate is constant at 3. Choice B correctly identifies the constant rate because the differences are equal, indicating a linear relationship. A common distractor like choice A might miscalculate the differences as varying, but actually they are consistently 3—double-check your subtractions to avoid this error. The three-method constant rate test: METHOD 1 (from table): Calculate Δy/Δx for each pair of consecutive points with equal Δx. All equal? Constant rate. Vary? Non-constant. METHOD 2 (from graph): Is it a straight line? Yes = constant rate. Curved? Non-constant. METHOD 3 (from formula): Is it y = mx + b form? Yes = constant rate m. Any other form (x², b^x, etc.)? Non-constant. Pick the method matching your representation! Don't confuse constant RATE with constant RATIO: constant rate (Δy/Δx equal) characterizes linear functions, constant ratio (y₂/y₁ equal) characterizes exponential functions. Check BOTH in a table: if differences are 3, 3, 3 → linear with rate 3. If ratios are 2, 2, 2 → exponential with base 2. If neither constant → some other type. Knowing which pattern to look for prevents confusing linear with exponential growth!

Question 19

Complete the square by using perfect square structure. Rewrite in the form (x+a)2+b(x+a)^2+b(x+a)2+b:

x2+8x+20x^2 + 8x + 20x2+8x+20

  1. (x+8)2−44(x+8)^2-44(x+8)2−44
  2. (x+4)2+4(x+4)^2+4(x+4)2+4 (correct answer)
  3. (x+4)2−4(x+4)^2-4(x+4)2−4
  4. (x−4)2+4(x-4)^2+4(x−4)2+4

Explanation: This question tests your ability to recognize mathematical patterns and structure in expressions—like difference of squares, sum and difference of cubes, or perfect square trinomials—and use those patterns to rewrite or factor the expression. Using structure means seeing an expression not just as a collection of terms, but as fitting a known pattern that enables transformation: completing the square transforms x² + 8x + 20 into a perfect square plus a constant! Let's complete the square for x² + 8x + 20: (1) take half the coefficient of x: 8/2 = 4, (2) square it: 4² = 16, (3) add and subtract this value: x² + 8x + 16 - 16 + 20, (4) recognize x² + 8x + 16 = (x + 4)² (perfect square trinomial!), (5) simplify: (x + 4)² - 16 + 20 = (x + 4)² + 4. Choice A correctly completes the square to get (x + 4)² + 4. Choice C has the right binomial (x + 4)² but incorrectly calculates the constant as -4 instead of +4—remember to combine -16 + 20 = +4, not just use -4! Completing the square recipe: for x² + bx + c, add and subtract (b/2)² to create the perfect square (x + b/2)², then simplify the constants. This structural transformation is crucial for solving quadratics, finding vertices of parabolas, and deriving the quadratic formula!

Question 20

A radioactive sample has a half-life of 12 years and starts with 80 grams. Create an exponential model for the remaining mass N(t)N(t)N(t) after ttt years, then determine how long it takes for the sample to decrease to 10 grams. Round to the nearest tenth of a year and interpret your result in context.

  1. Model: N(t)=80(12)t/12N(t)=80\left(\tfrac12\right)^{t/12}N(t)=80(21​)t/12. It reaches 10 g after t≈36.0t\approx 36.0t≈36.0 years. (correct answer)
  2. Model: N(t)=80(12)12tN(t)=80\left(\tfrac12\right)^{12t}N(t)=80(21​)12t. It reaches 10 g after t≈0.3t\approx 0.3t≈0.3 years.
  3. Model: N(t)=80(2)t/12N(t)=80(2)^{t/12}N(t)=80(2)t/12. It reaches 10 g after t≈36.0t\approx 36.0t≈36.0 years.
  4. Model: N(t)=80−12 tN(t)=80-\tfrac12\,tN(t)=80−21​t. It reaches 10 g after t≈140.0t\approx 140.0t≈140.0 years.

Explanation: This question tests your ability to create exponential models from real-world contexts (like compound interest, population growth, radioactive decay, or logarithmic scales) and solve them to answer practical questions. For half-life problems: if half-life is h, use model N(t) = N₀(0.5)^(t/h) where t/h is the number of half-lives that have passed. For this radioactive sample: (1) Initial mass N₀ = 80 g, half-life h = 12 years, (2) Model: N(t) = 80(0.5)^(t/12), (3) To find when N = 10, solve 10 = 80(0.5)^(t/12), (4) Divide: (0.5)^(t/12) = 0.125, (5) Note 0.125 = 1/8 = (0.5)³, so t/12 = 3, (6) Therefore t = 36 years. Choice A correctly creates the model N(t) = 80(0.5)^(t/12) and solves to get t ≈ 36.0 years, meaning the sample decays to 10 grams after 36 years (exactly 3 half-lives). Choice B incorrectly uses exponent 12t instead of t/12—the exponent should be the number of half-lives (t divided by h), not t times h! Half-life modeling strategy: (1) Use N(t) = N₀(0.5)^(t/h) where h is half-life, (2) The exponent t/h counts how many half-lives have passed, (3) After n half-lives, multiply initial amount by (0.5)^n, (4) To solve: recognize powers of 0.5 when possible (0.5¹ = 0.5, 0.5² = 0.25, 0.5³ = 0.125). Understanding that t/h counts half-lives makes these problems systematic!

Question 21

A streaming service has 5000 subscribers and grows by 12% each month.

Set up an exponential equation and solve for the number of months ttt until it reaches 10,000 subscribers.

Let ttt = time in months.

  1. Set up 5000(1.12)t=100005000(1.12)^t=100005000(1.12)t=10000. Then t=log⁡1.12(2)≈6.1t=\log_{1.12}(2)\approx 6.1t=log1.12​(2)≈6.1 months, so it takes about 6 months. (correct answer)
  2. Set up 5000(0.88)t=100005000(0.88)^t=100005000(0.88)t=10000. Then t=log⁡0.88(2)≈6.1t=\log_{0.88}(2)\approx 6.1t=log0.88​(2)≈6.1 months, so it takes about 6 months.
  3. Set up 5000+0.12t=100005000+0.12t=100005000+0.12t=10000. Then t=41666.7t=41666.7t=41666.7 months, so it takes about 41,667 months.
  4. Set up 5000(1.12)t=100005000(1.12)^t=100005000(1.12)t=10000. Then t=log⁡12(2)≈0.28t=\log_{12}(2)\approx 0.28t=log12​(2)≈0.28 months, so it takes about 0.28 months.

Explanation: This question tests your ability to translate complex real-world situations into mathematical equations (linear, quadratic, rational, or exponential) and solve them to answer practical questions. Exponential growth/decay problems arise when something grows or shrinks by a constant percent: 'grows 8% yearly' means multiply by 1.08 each year, giving formula amount = initial × (1.08)^t. To find when it reaches a specific value, set up equation like 500(1.08)^t = 1000 and solve using logarithms: (1.08)^t = 2, so t = ln(2)/ln(1.08) ≈ 9 years. The logarithm unlocks the exponent! For this subscriber growth, set up 5000(1.12)^t = 10000, simplify to (1.12)^t = 2, then t = ln(2)/ln(1.12) ≈ 6.1 months, meaning it takes about 6 months to double subscribers. Choice A correctly sets up the exponential growth equation and solves using logarithms to find approximately 6.1 months. A common error, as in choice C, models it linearly, leading to an absurdly long time frame. Context-to-equation type matching: (1) Constant rates and simple relationships → linear, (2) Area, projectile motion, optimization → quadratic, (3) Combined work rates, mixture concentrations, reciprocal relationships → rational, (4) Percent growth/decay over time → exponential. The context language tells you which type: 'per unit' suggests linear, 'area' suggests quadratic, 'together complete' suggests rational (adding reciprocals), 'percent per year' suggests exponential. Identify the type, then use the appropriate solving method! The reality check is essential: after solving, ask: (1) Does the answer satisfy the original equation? (substitute back), (2) Does it make sense in the real world? (no negative quantities, no fractional discrete items), (3) Have I answered what was asked? (find time, not rate; find width, not area). For rational equations, check that denominators aren't zero. For exponential, verify the value is reasonable for the time scale. This three-part check catches most errors and ensures your math answer is also a real-world answer!

Question 22

Identify the amplitude and period of the trigonometric function t(x)=2cos⁡(12x)t(x)=2\cos\left(\tfrac{1}{2}x\right)t(x)=2cos(21​x). (Relate to the parent function y=cos⁡xy=\cos xy=cosx.)

  1. Amplitude 222; period 4π4\pi4π (correct answer)
  2. Amplitude 12\tfrac{1}{2}21​; period 2π2\pi2π
  3. Amplitude 222; period π\piπ
  4. Amplitude 444; period 2π2\pi2π

Explanation: This question tests your ability to identify amplitude and period in trigonometric functions. Trigonometric functions like f(x) = a·sin(bx) + d have periodic (repeating) graphs with three key features: amplitude |a| is the vertical distance from the midline to a peak, period 2π/|b| is the horizontal length of one complete cycle, and midline y = d is the horizontal center line the graph oscillates around. The graph oscillates between y = d - |a| (minimum) and y = d + |a| (maximum), repeating this wave pattern every 2π/|b| units. For f(x) = 3sin(2x) + 1: amplitude 3, period π, midline y = 1, oscillating between -2 and 4. For t(x) = 2cos(½x), we identify: amplitude = |2| = 2 (the coefficient of cosine), and period = 2π/|½| = 2π/(½) = 2π · 2 = 4π (the ½ stretches the wave horizontally, making it take longer to complete one cycle). Choice A correctly identifies amplitude 2 and period 4π. Choice B incorrectly calculates amplitude as ½, Choice C has the wrong period, and Choice D misidentifies the amplitude. For trigonometric functions (sine and cosine): (1) Amplitude = |a| tells you how far from midline to peak (vertical stretch), (2) Period = 2π/|b| tells you how long one complete wave takes (horizontal compression if |b| > 1), (3) Midline y = d tells you the horizontal center (vertical shift). When |b| < 1 (like ½), the wave is stretched horizontally, increasing the period. Think of it as the wave moving in slow motion!

Question 23

For f(x)=(x−1)2f(x)=(x-1)^2f(x)=(x−1)2 with the restricted domain x≥1x\ge 1x≥1 (so the graph passes the horizontal line test on that domain), what is f−1(x)f^{-1}(x)f−1(x)?

  1. f−1(x)=1−xf^{-1}(x)=1-\sqrt{x}f−1(x)=1−x​
  2. f−1(x)=x−1f^{-1}(x)=\sqrt{x}-1f−1(x)=x​−1
  3. f−1(x)=1+xf^{-1}(x)=1+\sqrt{x}f−1(x)=1+x​ (correct answer)
  4. f−1(x)=±x+1f^{-1}(x)=\pm\sqrt{x}+1f−1(x)=±x​+1

Explanation: This question dives into finding inverses after restriction, building on why we restrict—excellent! For f(x) = (x-1)² restricted to x ≥ 1, it's one-to-one as it passes the horizontal line test on the right increasing branch from vertex at 1. To find the inverse: set y = (x-1)², swap to x = (y-1)², solve y = 1 + √x (positive root since y ≥ 1). The restriction to x ≥ 1 ensures we take the branch where the inverse outputs ≥1. Choice C correctly gives f⁻¹(x) = 1 + √x, matching the positive branch. Choice A uses minus, which would be for left restriction; D includes ±, but inverse must be single-valued. After restricting, swap and solve, choosing the branch per restriction— for ≥ vertex, it's plus. You're mastering how restrictions enable proper inverses—way to go!

Question 24

A theater’s revenue (in dollars) from selling nnn tickets is R(n)=15n.R(n)=15n.R(n)=15n. The theater has 250 seats and may sell out. What is an appropriate realistic domain for RRR?

  1. n∈{1,2,3,…,250}n\in\{1,2,3,\dots,250\}n∈{1,2,3,…,250}
  2. n∈{0,1,2,…,250}n\in\{0,1,2,\dots,250\}n∈{0,1,2,…,250} (correct answer)
  3. 0≤n≤2500\le n\le 2500≤n≤250 (all real numbers)
  4. All real numbers n≥0n\ge 0n≥0

Explanation: This question tests your understanding of how a function's domain relates both to its graph (which x-values have points) and to the real-world context (which values make sense practically). The domain is the set of allowed inputs, and for real-world functions, we distinguish: (1) mathematical domain (what the formula allows—like 'all reals' for polynomials), and (2) realistic domain (what makes sense in context—like 'positive integers' for number of items or 't≥0t \ge 0t≥0' for time). The realistic domain is usually more restrictive because real-world constraints (can't have negative time, can't produce -5 items, can't work 1.7 hours if discrete) limit what mathematical values are meaningful. Discrete vs continuous domains depend on what you're measuring: if counting distinct objects (people, tickets, items sold), the domain is discrete—only integers work, graph as separate dots. If measuring continuous quantities (time, distance, temperature, money as a continuous quantity), the domain is continuous—any value in an interval works, graph as connected line/curve. The physical nature of the quantity determines which! You can have 2.7 hours but not 2.7 people. In the theater revenue, nnn is tickets sold, discrete integers from 0 (no sales) to 250 (sold out). Choice B correctly identifies domain as n∈{0,1,2,…,250}n \in \{0,1,2,\dots,250\}n∈{0,1,2,…,250} based on whole ticket counts and seat limit. Choice C fails by excluding 0, but selling zero tickets is possible with zero revenue. Domain determination framework: (1) Start with mathematical domain—what does the formula allow? (polynomials: all reals; expression\sqrt{\text{expression}}expression​: expression ≥0\ge 0≥0; 1/expression1/\text{expression}1/expression: expression ≠0\ne 0=0; log⁡(expression)\log(\text{expression})log(expression): expression >0> 0>0), (2) Apply context constraints—can variable be negative? Are there upper limits? Must it be integer?, (3) Combine all restrictions—domain is where ALL conditions are met. Example: f(x)=x−3f(x) = \sqrt{x - 3}f(x)=x−3​ for measuring length has mathematical domain x≥3x \ge 3x≥3, and realistic domain also x≥3x \ge 3x≥3 (lengths are non-negative, and formula requires x≥3x \ge 3x≥3, so both agree). Discrete vs continuous quick-check: ask 'can there be in-between values?' If you're modeling number of students in a classroom, going from 20 to 21 students happens instantly (no 20.5 students exist)—discrete! If you're modeling temperature change from 20°C to 21°C, every value in between occurs—continuous! Context tells you: countable/distinct objects → discrete (dots on graph), measurable/varying quantities → continuous (line/curve on graph). This determines how you graph and what domain notation to use!

Question 25

The arithmetic sequence 4, 9, 14, 19,…4,\ 9,\ 14,\ 19,\dots4, 9, 14, 19,… defines values of a function at integer inputs with f(1)=4f(1)=4f(1)=4, f(2)=9f(2)=9f(2)=9, f(3)=14f(3)=14f(3)=14, etc. Write a linear function f(x)f(x)f(x) that extends this pattern.

  1. f(x)=4x+1f(x)=4x+1f(x)=4x+1
  2. f(x)=5x+1f(x)=5x+1f(x)=5x+1
  3. f(x)=4⋅5x−1f(x)=4\cdot 5^{x-1}f(x)=4⋅5x−1
  4. f(x)=5x−1f(x)=5x-1f(x)=5x−1 (correct answer)

Explanation: This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. To construct a linear function from two points, find the slope m = (y₂ - y₁)/(x₂ - x₁), then find the y-intercept b by substituting one point into y = mx + b and solving for b. Once you have m and b, you've got your function! For this arithmetic sequence with common difference 5, the linear function has slope m=5, and using f(1)=4: 4=51 + b gives b=-1, so f(x)=5x-1. Choice A correctly constructs the linear function with m=5 and b=-1 to extend the sequence. Choice B uses m=4, which doesn't match the difference—f(2)=42+1=9 ok but f(3)=12+1=13≠14; verify with multiple terms. Linear construction from two points recipe: (1) Find slope: m = (y₂ - y₁)/(x₂ - x₁), (2) Find y-intercept: substitute either point and m into y = mx + b, solve for b, (3) Write function: f(x) = [m]x + [b], (4) Verify: check that both original points work in your function. Example: (1, 4) and (3, 10) → m = (10-4)/(3-1) = 3, then 4 = 3(1) + b gives b = 1, so f(x) = 3x + 1. Check: f(1) = 4 ✓, f(3) = 10 ✓!