Algebra 2

Algebra 2 Practice Test: Practice Test 34

Practice Test 34 for Algebra 2: real questions and explanations from the Varsity Tutors practice-test pool.

0 / 25 answered

Question 1 of 25

Determine whether the system has one solution, no solution, or infinitely many solutions. If it has a solution, find it.

2x - y = 1 \\ 4x - 2y = 5 \end{cases}$$

Question Navigator

All questions

Question 1

Determine whether the system has one solution, no solution, or infinitely many solutions. If it has a solution, find it.

2x - y = 1 \\ 4x - 2y = 5 \end{cases}$$

No solution (correct answer)
Infinitely many solutions
Exactly one solution: $(2, 3)$
Exactly one solution: $(1, 1)$

Explanation: This question tests your ability to identify when a system has no solution, which occurs when the lines are parallel (same slope, different y-intercepts). A system of linear equations has no solution when the equations represent parallel lines that never intersect. Looking at 2x - y = 1 and 4x - 2y = 5, let's check if the second equation is a multiple of the first. If we multiply the first equation by 2, we get 4x - 2y = 2, but the second equation has 4x - 2y = 5. The left sides are identical but the right sides differ (2 ≠ 5), indicating parallel lines! Choice A correctly identifies this as no solution. We can verify by rewriting in slope-intercept form: First equation: y = 2x - 1. Second equation: y = 2x - 5/2. Both lines have slope 2 but different y-intercepts (-1 and -5/2), confirming they're parallel. The key to recognizing inconsistent systems: when you try to eliminate a variable and get a false statement like 0 = 3, or when the equations have proportional left sides but non-proportional right sides. Here, 4/2 = 2 and 2/1 = 2, but 5/1 ≠ 2, confirming no solution exists.

Question 2

A company models total cost as $C(q) = 1200 + 35q,$ where $q$ is the number of items produced. Which interpretation correctly identifies independent parts by chunking the expression into a fixed part and a part that depends on $q$ ?

Both 1200 and 35 depend on $q$ because they are in the same expression as $q$ .
The fixed part is $1200$ (does not depend on $q$ ), and the variable part is $35q$ (depends on $q$ ). (correct answer)
The fixed part is $35q$ because it is multiplied, and the variable part is 1200 because it is added.
The expression shows $q$ is independent because it is only in one term, so changing $q$ does not change $C(q)$ .

Explanation: This question tests your ability to interpret complicated expressions by viewing one or more parts as a single entity—a powerful technique for understanding structure and relationships in complex algebraic forms. When expressions get complex, chunking (viewing sub-expressions as single units) reveals structure: for C(q) = 1200 + 35q, we can view this as fixed costs plus variable costs, where each part plays a distinct role. In the expression C(q) = 1200 + 35q, chunking into meaningful parts shows: (1) 1200 is the fixed part—it doesn't contain q anywhere, so it's independent of production quantity, (2) 35q is the variable part—it directly depends on q through multiplication, (3) The sum structure means total cost equals fixed costs plus variable costs that scale with quantity. Choice B correctly identifies 1200 as the fixed part (doesn't depend on q) and 35q as the variable part (depends on q)—this reveals the standard linear cost model where you have overhead plus per-unit costs. Choice C backwards claims 35q is fixed and 1200 is variable, which makes no sense since 35q contains the variable q while 1200 is just a constant, and Choice D absurdly claims changing q doesn't change C(q), ignoring that q appears explicitly in the term 35q. Chunking strategy for linear models: (1) In expressions like a + bx, chunk into constant term (a) and variable term (bx), (2) The constant term is independent of the variable—it's the y-intercept or initial value, (3) The variable term shows rate of change—here, each additional item adds $35 to total cost. This structure has real meaning: 1200 represents fixed costs (rent, salaries) that you pay regardless of production, while 35q represents variable costs ($ 35 per item)—understanding this chunking helps interpret what happens as production changes!

Question 3

Two printers work at constant rates. Printer X can print a batch of flyers in 5 hours, and Printer Y can print the same batch in 10 hours. Write and solve an equation to find how long it takes to print one batch if both printers work together.

Let $t$ = time in hours to print one batch together.

Set up $\frac{1}{5}+\frac{1}{10}=t$ . Then $t=0.3$ hour. It takes 0.3 hour.
Set up $\frac{1}{5}+\frac{1}{10}=\frac{1}{t}$ . Then $t=\frac{10}{3}$ hours. It takes $\frac{10}{3}$ hours. (correct answer)
Set up $\frac{1}{5}-\frac{1}{10}=\frac{1}{t}$ . Then $t=10$ hours. It takes 10 hours.
Set up $\frac{5}{t}+\frac{10}{t}=1$ . Then $t=15$ hours. It takes 15 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. Printer X prints 1/5 of the batch per hour, Printer Y prints 1/10 per hour. Together they print 1/5 + 1/10 = 2/10 + 1/10 = 3/10 of the batch per hour. To find time for one whole batch: 1/5 + 1/10 = 1/t, which gives 3/10 = 1/t, so t = 10/3 hours (about 3.33 hours or 3 hours 20 minutes). Choice B correctly sets up the equation as 1/5 + 1/10 = 1/t, yielding t = 10/3 hours. Choice A incorrectly adds rates to get time directly (1/5 + 1/10 = t), which would mean the batch prints in 0.3 hours (18 minutes) - impossibly fast since even the faster printer alone takes 5 hours! 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 4

If $z_1$ and $z_2$ are the roots of $3x^2 - 12x + 15 = 0$ , what is the value of $|z_1|^2 + |z_2|^2$ ?

$8$
$16$
$12$
$10$ (correct answer)

Explanation: When you encounter a quadratic with complex roots and need to find expressions involving absolute values, think about using Vieta's formulas combined with properties of complex conjugates rather than solving for the individual roots. For the quadratic $3x^2 - 12x + 15 = 0$ , first check the discriminant: $b^2 - 4ac = (-12)^2 - 4(3)(15) = 144 - 180 = -36$ . Since this is negative, the roots are complex conjugates. Let's call them $z_1 = a + bi$ and $z_2 = a - bi$ . From Vieta's formulas: $z_1 + z_2 = \frac{12}{3} = 4$ and $z_1 \cdot z_2 = \frac{15}{3} = 5$ . Since the roots are conjugates, $z_1 + z_2 = (a + bi) + (a - bi) = 2a = 4$ , so $a = 2$ . Also, $z_1 \cdot z_2 = (a + bi)(a - bi) = a^2 + b^2 = 5$ , so $4 + b^2 = 5$ , giving us $b^2 = 1$ . Therefore: $|z_1|^2 = a^2 + b^2 = 4 + 1 = 5$ and $|z_2|^2 = a^2 + b^2 = 4 + 1 = 5$ . So $|z_1|^2 + |z_2|^2 = 5 + 5 = 10$ . Choice A ( $8$ ) might come from incorrectly using just $2a = 4$ . Choice B ( $16$ ) could result from squaring the sum of roots: $(z_1 + z_2)^2 = 16$ . Choice C ( $12$ ) might arise from adding the coefficients incorrectly. Remember: for complex conjugate roots of a quadratic, $|z_1|^2 + |z_2|^2$ always equals the product of the roots times 2, since both absolute values are equal.

Question 5

Consider the finite geometric series $\sum_{k=0}^{9} 4\left(\frac{1}{3}\right)^k.$ What is its exact value? Note: Since $\left|\frac{1}{3}\right|<1$ , the infinite geometric series would converge, but this question asks for the finite sum with $n=10$ terms.

$4\left(\frac{1}{3}\right)^{10}$
$\dfrac{4\left(1-\left(\frac{1}{3}\right)^9\right)}{1-\frac{1}{3}}$
$\dfrac{4\left(1-\left(\frac{1}{3}\right)^{10}\right)}{1-\frac{1}{3}}$ (correct answer)
$\dfrac{4\left(1-\left(\frac{1}{3}\right)^{10}\right)}{1+\frac{1}{3}}$

Explanation: This question tests your understanding of geometric series—the sum of terms from a geometric sequence—and applying $S_n = a(1 - r^n) / (1 - r)$ to summation notation with $|r|<1$ . The sum $\sum_{k=0}^9 4 (1/3)^k$ = $4 \sum_{k=0}^9 (1/3)^k$ , with inner sum $(1 - (1/3)^{10}) / (1 - 1/3)$ , n=10 terms. Rather than expanding, the formula handles it; note for infinite, it converges to $a/(1-r)$ , but here finite. Derivation's telescoping works for any n! Choice A correctly uses exponent 10 and denominator 1 - 1/3. Choice B has 9, perhaps off-by-one in indexing—k=0 to 9 is 10 terms. Confirm range (10 terms), simplify denominator (1 - 1/3 = 2/3), and compare to infinite for insight—this finite vs. infinite distinction is key, you're doing wonderfully!

Question 6

Write the system of linear equations as a single matrix equation $AX=B$ (use variable order $x, y$ ): [ \begin{cases} 3x+2y=8\ x-5y=-3 \end{cases} ]

$\begin{bmatrix}3&2\\1&-5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\-3\end{bmatrix}$ (correct answer)
$\begin{bmatrix}3&1\\2&-5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}8\\-3\end{bmatrix}$
$\begin{bmatrix}3&2\\1&-5\end{bmatrix}\begin{bmatrix}y\\x\end{bmatrix}=\begin{bmatrix}8\\-3\end{bmatrix}$
$\begin{bmatrix}3&2\\1&-5\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}-3\\8\end{bmatrix}$

Explanation: This question tests your ability to represent a system of linear equations as a single matrix equation in the form AX = B, where A is the coefficient matrix, X is the variable vector, and B is the constant vector. Matrix form AX = B is a compact way to write entire systems: the coefficient matrix A contains all the coefficients from the left sides of equations (organized by rows for equations and columns for variables), the variable vector X lists the unknowns as a column, and the constant vector B lists the right-hand side values. To convert the system 3x + 2y = 8 and x - 5y = -3 to matrix form: (1) Write coefficient matrix A by using equation 1 coefficients as row 1 and equation 2 coefficients as row 2, with columns for each variable in order: A = [[3, 2], [1, -5]] (first column is x-coefficients: 3, 1; second column is y-coefficients: 2, -5). (2) Write variable vector X = [x, y] as column (order matches columns of A). (3) Write constant vector B = [8, -3] as column (first equation's constant 8, second equation's constant -3, matching equation order). (4) Combine: [[3, 2], [1, -5]] times [x, y] = [8, -3]. Choice A correctly constructs the coefficient matrix A with proper row-column organization, variable vector X in correct order, and constant vector B matching the equations. Choice B has the rows and columns of matrix A transposed (swapped), putting coefficients in wrong positions - the second row should be [1, -5] not [2, -5], and the first column should contain x-coefficients [3, 1] not [3, 2]. Matrix equation construction recipe: (1) Label equations (first, second, third, etc.) and variables (x, y, z in consistent order), (2) Build matrix A: make rows from equations (row i = coefficients from equation i), make columns from variables (column j = coefficient of variable j in each equation), (3) Write X as column vector with variables in same order as A's columns, (4) Write B as column vector with constants in same order as equations. To verify your matrix equation represents the system correctly, perform the matrix multiplication AX mentally: the first row [3, 2] times [x, y] gives 3x + 2y = 8, and the second row [1, -5] times [x, y] gives x - 5y = -3, matching our original system perfectly!

Question 7

Use the identity $(a+b)^2=a^2+2ab+b^2$ to compute $101^2$ mentally by rewriting $101$ as $100+1$ . Which result is correct?

$101^2=10001$
$101^2=10201$ (correct answer)
$101^2=10101$
$101^2=10021$

Explanation: This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. The perfect square identity (a + b)² = a² + 2ab + b² allows us to compute squares mentally by breaking numbers into convenient parts. For 101², we write 101 = 100 + 1, so a = 100 and b = 1. Applying the identity: 101² = (100 + 1)² = 100² + 2(100)(1) + 1² = 10000 + 200 + 1 = 10201. Choice B correctly shows 101² = 10201. Choice A gives 10001, which misses the 2ab term entirely. Choice C gives 10101, which seems to add 101 instead of 201. Choice D gives 10021, transposing digits in the answer. To use perfect square identities for mental math: (1) Break the number into a round number plus/minus a small number, (2) Apply (a ± b)² = a² ± 2ab + b², (3) Calculate each piece mentally, (4) Add them up. For example, 99² = (100 - 1)² = 10000 - 200 + 1 = 9801. This technique makes seemingly difficult calculations manageable!

Question 8

Apply the sum of cubes identity to factor the polynomial $x^3+8$ completely over the integers.

$(x+2)(x^2+2x+4)$
$(x+2)(x^2-2x+4)$ (correct answer)
$(x-2)(x^2-2x+4)$
$(x^2+4)(x+2)$

Explanation: This question tests your understanding of polynomial identities—equations that are true for all values of the variables—and how to prove them algebraically or use them for applications like generating Pythagorean triples. A polynomial identity is different from a regular equation: it's true for ALL possible values of the variables, not just specific solutions. The sum of cubes identity a³ + b³ = (a + b)(a² - ab + b²) provides a direct way to factor expressions like x³ + 8. First, recognize that 8 = 2³, so we have x³ + 2³. Applying the identity with a = x and b = 2: x³ + 8 = x³ + 2³ = (x + 2)(x² - x(2) + 2²) = (x + 2)(x² - 2x + 4). Choice B correctly applies the sum of cubes pattern to factor x³ + 8 as (x + 2)(x² - 2x + 4). Choice A incorrectly has a plus sign in the middle term of the second factor (should be -2x, not +2x), choice C uses subtraction instead of addition in the first factor, and choice D doesn't follow the sum of cubes pattern at all. To apply sum or difference of cubes identities: (1) Recognize the form a³ ± b³, (2) Identify what a and b are (here x and 2), (3) Apply the pattern: sum gives (a + b)(a² - ab + b²), difference gives (a - b)(a² + ab + b²), (4) Substitute and simplify. Remember the sign patterns: sum of cubes has minus in the middle of the second factor, difference of cubes has plus!

Question 9

The table shows some values of function $p(x)$ . If $p$ has an inverse, which of the following could be the value of $p(6)$ ?

$1$
$3$
$5$
$9$ (correct answer)

Explanation: For $p$ to have an inverse, each output value must correspond to exactly one input value. The table shows outputs 1, 3, and 5 are already used, so $p(6)$ cannot equal any of these values. Only 9 is not already an output value in the table. Choices A, B, and C would create repeated output values, violating the requirement for an inverse function.

Question 10

Factor completely by recognizing repeated difference of squares structure:

$x^6 - 64$

$(x-4)(x^5+4x^4+16x^3+64x^2+256x+1024)$
$(x-2)(x^2+2x+4)(x+2)(x^2-2x+4)$ (correct answer)
$(x^3-8)(x^3+8)$
$(x^2-8)(x^4+8x^2+64)$

Explanation: This question tests your ability to recognize mathematical patterns and structure in expressions—specifically seeing $x^6 - 64$ as having multiple levels of structure that enable complete factorization. Using structure means recognizing $x^6 - 64 = (x^3)^2 - 8^2$ as a difference of squares, but also that $x^3 - 8$ is a difference of cubes! Let's factor $x^6 - 64$ completely: (1) see as $(x^3)^2 - 8^2$ (difference of squares), (2) factor: $(x^3 - 8)(x^3 + 8)$ , (3) recognize $x^3 - 8 = x^3 - 2^3$ (difference of cubes) and $x^3 + 8 = x^3 + 2^3$ (sum of cubes), (4) factor each: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ and $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$ , giving $(x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4)$ . Choice B correctly recognizes all the structure: first as difference of squares of cubes, then factoring each cube using the appropriate formula, resulting in the complete factorization $(x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4)$ . Choice A stops after the first step $(x^3 - 8)(x^3 + 8)$ without recognizing that both factors are cubes that can be factored further—when dealing with sixth powers, expect multiple levels of factoring! Multi-level structure recognition: $x^6$ often factors as either $(x^2)^3$ or $(x^3)^2$ , and here $(x^3)^2 - 8^2$ reveals difference of squares first. Then each resulting cube factors by sum/difference formulas. This problem beautifully demonstrates how different structural patterns (difference of squares, sum of cubes, difference of cubes) can all appear in one expression—recognizing each pattern in turn leads to complete factorization!

Question 11

A population model uses the annual growth expression $(1.08)^t$ , where $t$ is in years. Which equivalent expression best reveals the quarterly growth factor (4 quarters per year)?

$\left(1.08^{1/4}\right)^{4t}$ (correct answer)
$\left(1.08^{4}\right)^{t/4}$
$\left(1.08\right)^{t/4}$
$\left(\dfrac{1.08}{4}\right)^{4t}$

Explanation: This question tests your ability to use exponent properties to transform exponential expressions into equivalent forms that reveal different information, like converting annual growth rates to quarterly rates. The power-of-a-power property (b^a)^c = b^(ac) lets us rewrite expressions like (1.08)^t (8% annual growth) as ((1.08)^(1/4))^(4t) to reveal the quarterly growth rate: we break each year into 4 quarters and find the factor that, when applied 4 times (raised to power 4), gives the yearly factor 1.08. The quarterly factor is (1.08)^(1/4) ≈ 1.0194, meaning about 1.94% per quarter. Both forms equal the same value for any t, but one emphasizes annual compounding, the other quarterly! To convert annual expression (1.08)^t to quarterly form: (1) Recognize that t years equals 4t quarters, (2) Want form (something)^(4t), (3) That something is (1.08)^(1/4) because ((1.08)^(1/4))^(4t) = (1.08)^((1/4)×4t) = (1.08)^t by power-of-power property, (4) Calculate with calculator: (1.08)^(1/4) ≈ 1.0194, (5) So (1.08)^t = ((1.08)^(1/4))^(4t), revealing quarterly rate of about 1.94%. The exponent properties make the conversion systematic! Choice A correctly applies the power-of-a-power property to transform the expression and identifies the equivalent quarterly rate. Choice D divides the annual factor by 4 to get quarterly (1.08/4 = 0.27), but this gives SIMPLE interest division, not compound! For compound interest, the quarterly factor isn't found by division—we need (1.08)^(1/4), which is the 4th root of 1.08 ≈ 1.0194. Division would work for simple interest, but compound interest requires the fractional exponent! The power-of-a-power property (b^a)^c = b^(ac) is your main tool for time-base conversion: to convert annual rate factor b^t to quarterly, write as ((b)^(1/4))^(4t)—take 4th root of b for quarterly factor, raise to 4t (4 quarters times t years). To find a sub-period rate from annual: (1) Take annual factor (like 1.08 for 8%), (2) Raise to (1/n) where n is periods per year (1/4 for quarterly), (3) This gives per-period factor, (4) Subtract 1 and convert to percent for rate.

Question 12

A sequence is a function whose domain is a subset of the integers, so it is defined only at discrete inputs. Which set is the most appropriate domain for a sequence written as $a_1, a_2, a_3, \dots$ ?

All rational numbers $\mathbb{Q}$
All integers $\mathbb{Z}$
All real numbers $\mathbb{R}$
All positive integers $\{1,2,3,\dots\}$ (correct answer)

Explanation: This question tests your understanding that sequences are special functions where the domain is a subset of the integers—meaning they're only defined for whole number inputs like 1, 2, 3, not for values in between. A sequence is a function whose inputs are integers (often starting at 1 or 0) and whose outputs are the sequence terms: f(1) = first term, f(2) = second term, etc. Unlike continuous functions that have values for all real numbers, sequences have values only at integer positions. If you graphed a sequence, you'd see separate dots (like (1, a₁), (2, a₂), (3, a₃)), not a connected curve. This discrete nature makes sequences fundamentally different from functions like f(x) = x²! For a sequence like a₁, a₂, a₃, ..., the most appropriate domain is the positive integers {1,2,3,...}, as it matches the indexing starting from 1. Choice C correctly defines the integer domain as positive integers. A distractor like A (all reals) fails because sequences are discrete, not defined for non-integers—great job spotting that! Remember, sequences aren't continuous; their domain restricts to integers for that dotted graph effect. Keep exploring, and these concepts will click!

Question 13

Given the explicit formula $a_n = 18\left(\tfrac{2}{3}\right)^{n-1}$ for $n\ge 1$ , write an equivalent recursive definition for the sequence.

$a_1=18,\ a_{n+1}=a_n+\tfrac{2}{3}$
$a_1=18,\ a_{n+1}=\tfrac{2}{3}a_n$ (correct answer)
$a_1=12,\ a_{n+1}=\tfrac{2}{3}a_n$
$a_1=18,\ a_{n+1}=\tfrac{3}{2}a_n$

Explanation: This question tests your ability to write arithmetic and geometric sequences in both recursive form (each term from previous) and explicit form (any term directly from its position), and to translate between these forms. Translating between forms: if you have recursive (like a₁ = 3, aₙ₊₁ = aₙ + 5), extract a₁ = 3 and d = 5 (the amount added), then write explicit aₙ = 3 + 5(n - 1). If you have explicit (like aₙ = 2·3^(n-1)), extract a₁ = 2 (when n = 1) and r = 3 (the base), then write recursive a₁ = 2, aₙ₊₁ = 3aₙ. The parameters connect the two forms! The explicit aₙ=18·(2/3)^{n-1} gives a₁=18 and r=2/3 (the multiplier), so recursive is a₁=18, aₙ₊₁=(2/3)aₙ. Choice B correctly translates between forms, providing the equivalent recursive definition. Options like D use the reciprocal ratio, but checking a₂=12 confirms r=2/3—keep verifying terms to build confidence! Sequence type decision: calculate differences between consecutive terms (constant → arithmetic with that d) AND ratios of consecutive terms (constant → geometric with that r). For 5, 8, 11, 14: differences are 3, 3, 3 (arithmetic!), ratios are 8/5, 11/8, 14/11 (not constant). For 3, 6, 12, 24: differences are 3, 6, 12 (not constant), ratios are 2, 2, 2 (geometric!). This two-part check identifies the type reliably. Formula-writing checklist: (1) Identify type (arithmetic or geometric?), (2) Find first term a₁ (just look at the sequence), (3) Find d (subtract consecutive terms) or r (divide consecutive terms), (4) For recursive: state a₁ and write aₙ₊₁ = aₙ + d or aₙ₊₁ = r·aₙ, (5) For explicit: use aₙ = a₁ + (n-1)d or aₙ = a₁·r^(n-1). Follow these steps methodically and you'll get both forms correctly every time!

Question 14

A complex number $z=a+bi$ represents the point $(a,b)$ on the complex plane. Using the coordinate-geometry distance formula, the distance between $z_1=a+bi$ and $z_2=c+di$ is $|z_1-z_2|=\sqrt{(a-c)^2+(b-d)^2}$ , and the midpoint is $\dfrac{z_1+z_2}{2}$ . For $z_1=-2-i$ and $z_2=4+3i$ , what are the distance and the midpoint?

Distance $=\sqrt{52}$ , midpoint $=1+i$ (correct answer)
Distance $=\sqrt{40}$ , midpoint $=1+i$
Distance $=\sqrt{52}$ , midpoint $=2+\tfrac{1}{2}i$
Distance $=6+4i$ , midpoint $=1+i$

Explanation: This question tests your understanding of finding distances and midpoints between complex numbers on the complex plane using formulas that mirror 2D coordinate geometry. The distance between two complex numbers z₁ = a + bi and z₂ = c + di is the modulus of their difference: distance = |z₁ - z₂| = √((a - c)² + (b - d)²), which is exactly the 2D distance formula! The midpoint formula: midpoint = (z₁ + z₂)/2 = ((a + c)/2) + ((b + d)/2)i, averaging the real and imaginary parts separately. For z₁ = -2 - i and z₂ = 4 + 3i: (1) Calculate difference: (-2 - i) - (4 + 3i) = -6 - 4i. (2) Find modulus: |−6 - 4i| = √((-6)² + (-4)²) = √(36 + 16) = √52. (3) For midpoint: (-2 - i) + (4 + 3i) = 2 + 2i, then divide by 2 to get 1 + i. Choice A correctly applies both formulas to get distance = √52 and midpoint = 1 + i. Choice B incorrectly calculates the distance as √40 instead of √52, perhaps making an error when squaring -6 or -4, or when adding 36 + 16. Always double-check your arithmetic: (-6)² = 36 and (-4)² = 16, so 36 + 16 = 52, not 40! The beauty of these formulas is that they connect complex number arithmetic with geometry seamlessly—the complex plane truly is a coordinate system!

Question 15

A manufacturing company plans to increase production by 8% each quarter for the next 5 quarters, starting with an initial production of 2,400 units in Quarter 1. What is the total number of units produced over all 5 quarters?

$14,112$ units (correct answer)
$15,369$ units
$13,056$ units
$14,598$ units

Explanation: This is a finite geometric series with first term $a = 2400$ , common ratio $r = 1.08$ , and $n = 5$ terms. Using the formula $S_n = a \cdot \frac{1-r^n}{1-r}$ : $S_5 = 2400 \cdot \frac{1-(1.08)^5}{1-1.08} = 2400 \cdot \frac{1-1.4693}{-0.08} = 2400 \cdot \frac{-0.4693}{-0.08} = 2400 \cdot 5.8663 = 14,112$ units. Choice B uses $r = 0.08$ instead of $1.08$ . Choice C calculates only the growth amount, not total production. Choice D incorrectly uses simple interest formula.

Question 16

Solve and evaluate using technology: $12\cdot 10^{t/4}=500$ . Give $t$ in exact logarithmic form and as a decimal approximation.

$t=4\log\left(\frac{500}{12}\right)\approx 6.48$ (correct answer)
$t=\dfrac{\log\left(\frac{500}{12}\right)}{4}\approx 0.405$
$t=4\log\left(\frac{12}{500}\right)\approx -6.48$
$t=4\ln\left(\frac{500}{12}\right)\approx 14.92$

Explanation: This question tests your ability to solve exponential equations by taking logarithms of both sides and using the inverse relationship to isolate the variable. When solving exponential equations like 12·10^(t/4) = 500 where the exponent contains t/4, logarithms unlock the solution: first isolate the exponential by dividing both sides by 12 to get 10^(t/4) = 500/12 ≈ 41.667, then take log base 10 of both sides to get log(10^(t/4)) = log(500/12), and using the inverse property log(10^(t/4)) = t/4, we get t/4 = log(500/12), so t = 4·log(500/12). To solve 12·10^(t/4) = 500 systematically: divide by 12 to get 10^(t/4) = 41.667, take common log of both sides: log(10^(t/4)) = log(41.667), apply the inverse property to get t/4 = log(41.667) ≈ 1.620, and multiply by 4: t = 4·log(41.667) = 4·log(500/12) ≈ 4·1.620 ≈ 6.48. Choice A correctly isolates the exponential, takes the common logarithm, and multiplies by 4 to get t = 4·log(500/12) ≈ 6.48. Choice D uses natural log (ln) instead of common log, giving t = 4·ln(500/12) ≈ 4·3.730 ≈ 14.92, which is wrong because we need log base 10 to match our base 10 exponential—using ln would require additional conversion. Isolation before logarithms: ALWAYS isolate the exponential expression 10^(t/4) before taking logarithms. If you have 12·10^(t/4) = 500, first divide by 12 to get 10^(t/4) = 41.667, THEN take log. When the variable appears as t/4 in the exponent, remember that after taking the logarithm you get t/4 = log(41.667), so you must multiply by 4 to solve for t—don't divide by 4!

Question 17

Verify that the functions $f(x)=4x-7$ and $g(x)=\dfrac{x+7}{4}$ are inverses by composition. Compute both $f(g(x))$ and $g(f(x))$ and determine whether each equals $x$ .

$f(g(x))=4\left(\dfrac{x+7}{4}\right)-7=x$ and $g(f(x))=\dfrac{(4x-7)+7}{4}=x$ , so they are inverses. (correct answer)
$f(g(x))=4\left(\dfrac{x+7}{4}\right)-7=x$ but $g(f(x))=\dfrac{4x-7}{4}=x-\dfrac{7}{4}$ , so they are not inverses.
$f(g(x))=4\left(\dfrac{x+7}{4}\right)-7=x+7$ and $g(f(x))=\dfrac{(4x-7)+7}{4}=x$ , so they are not inverses.
$f(g(x))=\dfrac{x+7}{4}-7=\dfrac{x-21}{4}$ and $g(f(x))=\dfrac{x+7}{4}$ , so they are inverses.

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! Let's compute: f(g(x)) = 4*((x+7)/4) - 7 = (x + 7) - 7 = x, and g(f(x)) = (4x - 7 + 7)/4 = 4x/4 = x, so both simplify to x. Choice A correctly verifies both compositions equal x and determines they are inverses. A common distractor like choice B might miscompute one composition, such as forgetting to add 7 in g(f(x)), leading to an incorrect conclusion that they are not inverses. 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! Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

Question 18

A savings account starts with $200$ dollars and increases by $10\%$ each month. Let $t$ be the number of months since the start. Write an exponential function $S(t)$ that models the balance.

$S(t)=200+0.1t$
$S(t)=200(0.9)^t$
$S(t)=200(1.1)^t$ (correct answer)
$S(t)=210(1.1)^t$

Explanation: This question tests your ability to construct linear or exponential functions from given information like points, tables, graphs, or descriptions of relationships. For exponential functions, find the initial value a (the y-value when x = 0, or work backward if needed) and the growth/decay factor b (divide consecutive y-values: b = y₂/y₁ when x increases by 1). Then write f(x) = a·b^x. 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! With initial $200 and 10% monthly growth, a=200 and b=1.1, so S(t)=200*(1.1)^t. Choice A correctly constructs the exponential function with initial value 200 and growth factor 1.1 from the description. A linear distractor like choice D might add percentages additively—remember exponential models multiplicative growth! Exponential construction strategy: (1) Find initial value: if you have x = 0 in data, that y is your a; otherwise calculate backward using the pattern, (2) Find base: divide consecutive y-values (with x differing by 1): b = y_{x+1}/y_x—should be constant for exponential, (3) Write f(x) = a·b^x, (4) Verify with all data points. Example: points (0, 100) and (1, 110) and (2, 121) → a = 100, b = 110/100 = 1.1 (check: 121/110 = 1.1 ✓), so f(x) = 100·(1.1)^x. The ratio test both identifies the type and gives you the base!

Question 19

A company models the average cost per item (in dollars) when producing $n$ items by $A(n)=\dfrac{500}{n}+12$ . In context, $n$ is the number of items produced in a batch, and the factory can produce at most 200 items per batch. What is an appropriate realistic domain for $n$ ?

All real numbers except $n=0$
$n\in[0,200]$
$n\in(0,200]$
$n\in\{1,2,3,\dots,200\}$ (correct answer)

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). 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. Here, $n$ represents the number of items produced, which must be positive integers (can't produce a fraction or zero items, as average cost is undefined at $n=0$ and doesn't make sense contextually), up to 200 per batch. Choice B correctly identifies the domain as $\{1,2,3,\dots,200\}$ based on the discrete nature of countable items and the factory's maximum. A distractor like Choice A fails by treating $n$ as continuous and excluding zero properly but allowing non-integers, which isn't realistic for whole items. 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 $20^\circ\text{C}$ to $21^\circ\text{C}$ , 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! Excellent work—you're getting great at distinguishing these concepts!

Question 20

If the remainder when $p(x)=x^3-2x^2+kx-4$ is divided by $(x-2)$ is $6$ , what is the value of $k$ ? (Use remainder $=p(2)$ .)

$k=5$ (correct answer)
$k=-1$
$k=3$
$k=1$

Explanation: This question tests your understanding of the Remainder Theorem—a powerful shortcut that lets you find the remainder when dividing a polynomial by (x - a) without doing any long division! The Remainder Theorem states that when you divide a polynomial p(x) by (x - a), the remainder is simply p(a)—just substitute a into the polynomial and evaluate! This works because of how polynomial division works: p(x) = (x - a)·q(x) + r, where r is the remainder. Substituting x = a: p(a) = (a - a)·q(a) + r = 0 + r = r. So the remainder r equals p(a)—brilliant! Here, divisor (x - 2), a = 2, and remainder = 6, so p(2) = 6: (2)^3 - 2*(2)^2 + k*(2) - 4 = 8 - 2*4 + 2k - 4 = 8 - 8 + 2k - 4 = 2k - 4 = 6, so 2k = 10, k = 5. Choice D correctly finds k = 5. Choice A says k = -1, perhaps from setting 2k - 4 = 0 instead of 6 or a solving error like 2k = 2, but always solve the equation carefully! Using the Remainder Theorem: (1) Identify the divisor (x - a) and extract a (remember: (x + 3) = (x - (-3)), so a = -3), (2) Substitute a for every x in p(x), (3) Calculate carefully (use parentheses for negative values!), (4) That result is your remainder. If it equals 0, (x - a) is a factor! This method is dramatically faster than polynomial long division.

Question 21

Check if $f(x)=\dfrac{x+3}{2}$ and $g(x)=2x+3$ are inverse functions by computing both $f(g(x))$ and $g(f(x))$ and comparing each to $x$ .

$f(g(x))=\dfrac{(2x+3)+3}{2}=x+3$ and $g(f(x))=2\left(\dfrac{x+3}{2}\right)+3=x+6$ , so they are not inverses. (correct answer)
$f(g(x))=\dfrac{2x+3}{2}+3=x+\dfrac{9}{2}$ and $g(f(x))=\dfrac{x+3}{2}+3=\dfrac{x+9}{2}$ , so they are inverses.
$f(g(x))=\dfrac{(2x+3)+3}{2}=x+3$ and $g(f(x))=2\left(\dfrac{x+3}{2}\right)+3=x+6$ , so they are inverses because both compositions are linear.
$f(g(x))=\dfrac{(2x+3)+3}{2}=x$ and $g(f(x))=2\left(\dfrac{x+3}{2}\right)+3=x$ , so they are inverses.

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. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! Here, f(g(x)) = [ (2x + 3) + 3 ] / 2 = (2x + 6)/2 = x + 3 ≠ x, and g(f(x)) = 2[ (x + 3)/2 ] + 3 = (x + 3) + 3 = x + 6 ≠ x, so neither equals x. Choice A correctly determines they are not inverses by showing both compositions. A distractor like choice D fails by mistakenly simplifying to x, perhaps by subtracting instead of adding 3. Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence! Don't worry if they weren't inverses—this helps you spot non-inverses too!

Question 22

Use technology to solve the system by the inverse-matrix method. Write $AX=B$ and compute $X=A^{-1}B$ .

$\begin{cases} 2x+y+z=4 \\ x+3y+2z=5 \\ 3x+y+4z=6 \end{cases}$

What is $X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ ?

$\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$
$\begin{bmatrix} \tfrac{7}{6} \\ \tfrac{5}{6} \\ \tfrac{5}{6} \end{bmatrix}$
$\begin{bmatrix} \tfrac{5}{6} \\ \tfrac{7}{6} \\ \tfrac{5}{6} \end{bmatrix}$
$\begin{bmatrix} \tfrac{5}{6} \\ \tfrac{5}{6} \\ \tfrac{7}{6} \end{bmatrix}$ (correct answer)

Explanation: This question tests your understanding of using matrix inverses to solve linear systems—a powerful method that reduces solving a system to a single matrix multiplication once the inverse is found. For the 3×3 system given, we write it as $AX = B$ where $A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 3 & 2 \\ 3 & 1 & 4 \end{bmatrix}$ , $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ , and $B = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}$ . Using technology to find $A^{-1}$ and compute $X = A^{-1}B$ , we obtain the solution vector $X = \begin{bmatrix} \frac{5}{6} \\ \frac{5}{6} \\ \frac{7}{6} \end{bmatrix}$ , which corresponds to choice D. Choices B and C show different arrangements of these fraction values, likely from misreading the solution vector order, while choice A suggests all variables equal 1, which doesn't satisfy the system. The matrix inverse method is particularly powerful for systems with fractional solutions—technology handles the arithmetic precisely where hand calculation might introduce rounding errors. For 3×3 systems, always use technology as the standard recommends—it ensures accuracy and saves tremendous time!

Question 23

Rewrite $\frac{3x^2-2x+4}{x-1}$ in the form $q(x)+\frac{r(x)}{x-1}$ , where $\deg(r)<\deg(x-1)$ . (You may use inspection or long division.)

$3x+1+\dfrac{5}{x-1}$ (correct answer)
$3x+2+\dfrac{4}{x-1}$
$3x+1+\dfrac{5x}{x-1}$
$3x-1+\dfrac{5}{x-1}$

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. For $3x^2 - 2x + 4$ divided by $x - 1$ , division gives $3x + 1$ with remainder 5, verifiable by inspection or synthetic division. Choice A correctly provides the quotient $3x + 1$ and remainder 5, with $\deg(r) < 1$ . Choice B might result from a sign error in subtraction, yielding -1 instead of +1. The degree requirement tells you to stop when the remainder is a constant for a linear divisor, preventing unnecessary further steps.

Question 24

A radioactive sample has a half-life of 12 years and initially has a mass of 160 grams. Create an exponential model for the remaining mass $N(t)$ after $t$ years, then solve for when 20 grams remain. Round your answer to the nearest tenth of a year and interpret your result.

Model: $N(t)=160\left(\tfrac12\right)^{t/12}$ . It reaches 20 g after about $t\approx 36.0$ years. (correct answer)
Model: $N(t)=160\left(\tfrac12\right)^{12t}$ . It reaches 20 g after about $t\approx 0.3$ years.
Model: $N(t)=20\left(\tfrac12\right)^{t/12}$ . It reaches 20 g after about $t\approx 36.0$ years.
Model: $N(t)=160(0.5)t/12$ . It reaches 20 g after about $t\approx 10.7$ years.

Explanation: This question tests your ability to create exponential models for radioactive decay using half-life and solve for when a specific amount remains. For half-life problems, use N(t) = N₀(1/2)^(t/h) where N₀ is initial amount, h is half-life, and t/h represents the number of half-lives elapsed. With N₀ = 160 g and h = 12 years: N(t) = 160(1/2)^(t/12). To find when N = 20: set 20 = 160(1/2)^(t/12), divide by 160 to get 1/8 = (1/2)^(t/12), recognize 1/8 = (1/2)³, so (1/2)^(t/12) = (1/2)³, therefore t/12 = 3, giving t = 36 years. Choice A correctly models decay with the half-life formula and finds that 20 g remains after 36 years (exactly 3 half-lives). Choice B incorrectly uses (1/2)^(12t) instead of (1/2)^(t/12), which would mean 12 half-lives per year instead of one half-life per 12 years—the exponent should be t divided by the half-life period! Understanding that t/h counts half-lives makes these problems systematic: from 160 to 20 is 160→80→40→20, which is 3 halvings.

Question 25

A function is sampled at integer inputs, producing the sequence $7,\ 12,\ 17,\ 22$ . Over equal input intervals of $1$ , which statement correctly identifies the growth pattern and the most likely function type?

Because the terms increase, it must be exponential growth (all increasing sequences are exponential).
The ratios are constant ( $\tfrac{12}{7}=\tfrac{17}{12}=\tfrac{22}{17}$ ), so it matches exponential growth and corresponds to a geometric sequence.
The differences are $5,5,5$ , so it matches exponential growth because exponentials add the same amount each step.
The differences are constant ( $+5$ each step), so it matches linear growth and corresponds to an arithmetic sequence. (correct answer)

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). Given the sequence 7, 12, 17, 22, let's check both patterns. Differences: 12 - 7 = 5, 17 - 12 = 5, 22 - 17 = 5. All differences equal 5 - constant! Ratios: 12/7 ≈ 1.71, 17/12 ≈ 1.42, 22/17 ≈ 1.29. The ratios are different - not constant! Choice A correctly identifies constant differences of +5, indicating linear growth and an arithmetic sequence. This suggests the function is f(x) = 5x + 7. Choice B incorrectly claims the ratios are constant (they're not equal: 12/7 ≠ 17/12 ≠ 22/17). Choice C makes the false claim that all increasing sequences are exponential - many increasing sequences are linear! Choice D correctly identifies the differences but wrongly associates constant differences with exponential growth instead of linear growth. To determine if a function is linear or exponential from a table: (1) calculate differences between consecutive y-values, (2) calculate ratios between consecutive y-values, (3) if differences are constant → linear, if ratios are constant → exponential. Why this matters: the growth pattern reveals the function type and lets you predict the next value - here, add 5 to get 27!

Opening subject page...