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ACT Math

ACT Math Practice Test: Practice Test 18

Practice Test 18 for ACT Math: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

Convert ( \frac{2}{3} ) to a percent.

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Question 1

Convert ( \frac{2}{3} ) to a percent.

  1. 65%
  2. 60%
  3. 66.67% (correct answer)
  4. 70%

Explanation: This question asks to convert the fraction 2/3 to a percent. To convert a fraction to a percent, divide the numerator by the denominator and multiply by 100. 2 ÷ 3 = 0.6667..., and 0.6667 × 100 = 66.67%. Choice B incorrectly calculated 60%, which would be 3/5, not 2/3.

Question 2

What is 3u−v3\mathbf{u} - \mathbf{v}3u−v if u=⟨2,4⟩\mathbf{u} = \langle 2, 4 \rangleu=⟨2,4⟩ and v=⟨1,−1⟩\mathbf{v} = \langle 1, -1 \ranglev=⟨1,−1⟩?

  1. ⟨7,11⟩\langle 7, 11 \rangle⟨7,11⟩
  2. ⟨6,13⟩\langle 6, 13 \rangle⟨6,13⟩
  3. ⟨5,13⟩\langle 5, 13 \rangle⟨5,13⟩ (correct answer)
  4. ⟨5,11⟩\langle 5, 11 \rangle⟨5,11⟩

Explanation: This problem involves a linear combination of vectors. We compute 3u−v=3⟨2,4⟩−⟨1,−1⟩3\mathbf{u} - \mathbf{v} = 3\langle 2, 4 \rangle - \langle 1, -1 \rangle3u−v=3⟨2,4⟩−⟨1,−1⟩. First multiply: 3⟨2,4⟩=⟨6,12⟩3\langle 2, 4 \rangle = \langle 6, 12 \rangle3⟨2,4⟩=⟨6,12⟩. Then subtract: ⟨6,12⟩−⟨1,−1⟩=⟨6−1,12−(−1)⟩=⟨5,13⟩\langle 6, 12 \rangle - \langle 1, -1 \rangle = \langle 6 - 1, 12 - (-1) \rangle = \langle 5, 13 \rangle⟨6,12⟩−⟨1,−1⟩=⟨6−1,12−(−1)⟩=⟨5,13⟩. Apply scalar multiplication first, then vector subtraction.

Question 3

A function is defined by f(x)=3x−7f(x)=3x-7f(x)=3x−7. What is f(5)f(5)f(5)?

Use substitution: f(x)=3x−7f(x)=3x-7f(x)=3x−7, so f(5)=3(5)−7f(5)=3(5)-7f(5)=3(5)−7.

  1. 888 (correct answer)
  2. −22-22−22
  3. 222222
  4. −8-8−8

Explanation: We need to find f(5) where f(x) = 3x - 7. To find f(5), we substitute x = 5 into the function: f(5) = 3(5) - 7 = 15 - 7 = 8. The answer is 8.

Question 4

A bag contains 6 red and 4 yellow balls. What is the probability of drawing two red balls in a row without replacement?

  1. 13\frac{1}{3}31​ (correct answer)
  2. 12\frac{1}{2}21​
  3. 14\frac{1}{4}41​
  4. 16\frac{1}{6}61​

Explanation: The bag contains 6 red + 4 yellow = 101010 total balls. For drawing two red balls without replacement, we need P(1st red) × P(2nd red|1st red). The first draw has 6 red out of 10 total, so P(1st red) = 610\frac{6}{10}106​. After removing one red ball, there are 5 red out of 9 remaining, so P(2nd red|1st red) = 59\frac{5}{9}95​. Therefore P(two red) = 610\frac{6}{10}106​ × 59\frac{5}{9}95​ = 3090\frac{30}{90}9030​ = 13\frac{1}{3}31​. Choice C (1/4) might incorrectly assume replacement.

Question 5

Factor: x2−6x+9x^2 - 6x + 9x2−6x+9

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

Explanation: To factor x2−6x+9x^2 - 6x + 9x2−6x+9, we recognize this as a perfect square trinomial. We need two numbers that multiply to 9 and add to -6: (-3)(-3) = 9 and (-3) + (-3) = -6. This gives us (x−3)(x−3)=(x−3)2(x - 3)(x - 3) = (x - 3)^2(x−3)(x−3)=(x−3)2. Perfect square trinomials follow the pattern a2−2ab+b2=(a−b)2a^2 - 2ab + b^2 = (a - b)^2a2−2ab+b2=(a−b)2.

Question 6

If △XYZ∼△ABC\triangle XYZ \thicksim \triangle ABC△XYZ∼△ABC with XY=5XY = 5XY=5 and AB=15AB = 15AB=15, what is the ratio of YZYZYZ to BCBCBC?

  1. 1:3 (correct answer)
  2. 3:1
  3. 1:1
  4. 2:3

Explanation: Triangles XYZ and ABC are similar, meaning corresponding sides are proportional. The sides XY and AB correspond to each other, with XY=5XY = 5XY=5 and AB=15AB = 15AB=15. The scale factor from triangle XYZ to triangle ABC is AB/XY=15/5=3/1AB/XY = 15/5 = 3/1AB/XY=15/5=3/1. Since the triangles are similar, all corresponding sides have the same ratio. Therefore, YZ corresponds to BC, and the ratio YZ:BC=1:3YZ:BC = 1:3YZ:BC=1:3.

Question 7

What angle θ\thetaθ (in degrees) has sin⁡(θ)=22\sin(\theta)=\frac{\sqrt{2}}{2}sin(θ)=22​​, where 0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ0∘≤θ≤90∘?​​

  1. 30∘30^\circ30∘
  2. 45∘45^\circ45∘ (correct answer)
  3. 60∘60^\circ60∘
  4. 90∘90^\circ90∘

Explanation: We need to find θ where sin(θ) = √2/2. This is a standard unit circle value: sin(45°) = √2/2. We can verify: in a 45-45-90 triangle with hypotenuse √2, both legs equal 1, so sin(45°) = 1/√2 = √2/2.

Question 8

A matrix AAA is used to encode a 2-variable system. If A=[0−321],A=\begin{bmatrix}0 & -3\\ 2 & 1\end{bmatrix},A=[02​−31​], what is −2A-2A−2A?

  1. [0−1−4−2]\begin{bmatrix}0 & -1\\ -4 & -2\end{bmatrix}[0−4​−1−2​]
  2. [0−6−4−2]\begin{bmatrix}0 & -6\\ -4 & -2\end{bmatrix}[0−4​−6−2​]
  3. [0642]\begin{bmatrix}0 & 6\\ 4 & 2\end{bmatrix}[04​62​]
  4. [06−4−2]\begin{bmatrix}0 & 6\\ -4 & -2\end{bmatrix}[0−4​6−2​] (correct answer)

Explanation: The operation is scalar multiplication by -2, multiplying every entry of the matrix by this scalar. For -2A, (1,1): -20 = 0; (1,2): -2(-3) = 6; (2,1): -22 = -4; (2,2): -21 = -2. This reflects and scales the encoding matrix uniformly. The result is the matrix [[0, 6], [-4, -2]]. Choice B flips the sign of the (1,2) entry, possibly from mishandling the negative scalar.

Question 9

A runner's distance from the starting line increases at a constant rate. The relationship is modeled by y=0.25xy = 0.25xy=0.25x, where xxx is time in seconds and yyy is distance in meters. What is the meaning of the slope in this context?

  1. The runner starts 0.25 meters ahead of the starting line.
  2. The runner runs 0.25 meters per second. (correct answer)
  3. The runner runs 4 seconds per meter.
  4. The runner runs 0.25 meters in total.

Explanation: The model y = 0.25x represents distance (y in meters) versus time (x in seconds) for a runner. The slope 0.25 means the runner's distance increases by 0.25 meters for each second that passes - this is the runner's speed of 0.25 meters per second. There is no y-intercept term, meaning the runner starts at the starting line (0 meters when x = 0). Choice C incorrectly inverts the units to seconds per meter. Choice A misinterprets the slope as a starting position.

Question 10

Let A=[−213−1]A=\begin{bmatrix}-2 & 1\\ 3 & -1\end{bmatrix}A=[−23​1−1​] and B=[50−42].B=\begin{bmatrix}5 & 0\\ -4 & 2\end{bmatrix}.B=[5−4​02​]. Which of the following is the product ABABAB?

  1. [−14192−2]\begin{bmatrix}-14 & 19\\ 2 & -2\end{bmatrix}[−142​19−2​]
  2. [−100−12−2]\begin{bmatrix}-10 & 0\\ -12 & -2\end{bmatrix}[−10−12​0−2​]
  3. [−14211−2]\begin{bmatrix}-14 & 2\\ 11 & -2\end{bmatrix}[−1411​2−2​]
  4. [−14219−2]\begin{bmatrix}-14 & 2\\ 19 & -2\end{bmatrix}[−1419​2−2​] (correct answer)

Explanation: The operation is matrix multiplication, requiring dot products of rows from A with columns from B. For AB, (1,1) is −2∗5+1∗(−4)=−14-2*5 + 1*(-4) = -14−2∗5+1∗(−4)=−14; (1,2) is −2∗0+1∗2=2-2*0 + 1*2 = 2−2∗0+1∗2=2; (2,1) is 3∗5+(−1)∗(−4)=193*5 + (-1)*(-4) = 193∗5+(−1)∗(−4)=19; (2,2) is 3∗0+(−1)∗2=−23*0 + (-1)*2 = -23∗0+(−1)∗2=−2. Ensure you multiply and add in the correct order for each entry. The result is the matrix [−14219−2]\begin{bmatrix} -14 & 2 \\ 19 & -2 \end{bmatrix}[−1419​2−2​]. Choice C uses 11 instead of 19, likely from forgetting to add 4 in the (2,1) calculation.

Question 11

Tickets to a high school play cost 12foradultsand12 for adults and 12foradultsand8 for children. On opening night, 150 tickets were sold and total revenue was $1,440. How many more children's tickets were sold than adult tickets?

  1. 303030 (correct answer)
  2. 454545
  3. 606060
  4. 909090

Explanation: This is a systems of equations word problem testing multi-step algebraic modeling. Choice A (30) is correct — set up: let A = adult tickets and C = children's tickets. Two equations: A + C = 150 (total tickets) and 12A + 8C = 1,440 (total revenue). Solve by substitution: C = 150 − A → 12A + 8(150 − A) = 1,440 → 12A + 1,200 − 8A = 1,440 → 4A = 240 → A = 60. Then C = 150 − 60 = 90. Difference: C − A = 90 − 60 = 30. Choice B (45) likely comes from an arithmetic error mid-solve, perhaps computing 4A = 180 → A = 45. Choice C (60) reports the number of adult tickets — finding A but not completing the final step (finding the difference). Choice D (90) reports the number of children's tickets — finding C but not subtracting A. Pro tip: Systems word problems require a final step after solving for variables. Read the question again carefully — here it asks "how many MORE children's tickets," which means C − A, not just C or A alone. After solving the system, always return to the original question to make sure you're reporting the right quantity.

Question 12

A car rental price is modeled by a line with slope −2-2−2 that passes through the point (3,5)(3,5)(3,5). Which equation represents this line?

  1. y=−2x+11y=-2x+11y=−2x+11 (correct answer)
  2. y=2x−1y=2x-1y=2x−1
  3. y=−2x−1y=-2x-1y=−2x−1
  4. y=−12x+132y=-\dfrac{1}{2}x+\dfrac{13}{2}y=−21​x+213​

Explanation: We need the equation of a line with slope -2 passing through point (3,5). Using point-slope form: y - 5 = -2(x - 3). Expanding: y - 5 = -2x + 6, so y = -2x + 11. The answer is A: y = -2x + 11. Choice C (y = -2x - 1) has the correct slope but calculates the y-intercept incorrectly.

Question 13

A worker earns $15.00 per hour for the first 40 hours worked in a week. For any hours worked over 40, the worker earns $22.50 per hour. If the worker works 46 hours in one week, what is the total amount earned?

  1. $690.00
  2. $735.00 (correct answer)
  3. $750.00
  4. $1{,}035.00

Explanation: This is an arithmetic and rates question testing overtime pay calculations. Choice B (735.00)iscorrect—regularpayforthefirst40hours:40×735.00) is correct — regular pay for the first 40 hours: 40 × 735.00)iscorrect—regularpayforthefirst40hours:40×15.00 = 600.00.Overtimehours:46−40=6hours.Overtimepay:6×600.00. Overtime hours: 46 − 40 = 6 hours. Overtime pay: 6 × 600.00.Overtimehours:46−40=6hours.Overtimepay:6×22.50 = 135.00.Total:135.00. Total: 135.00.Total:600 + 135=135 = 135=735.00. Choice A (690.00)appliestheregular690.00) applies the regular 690.00)appliestheregular15.00 rate to all 46 hours, ignoring the overtime premium entirely: 46 × 15=15 = 15=690. Choice C (750.00)resultsfromasmallarithmeticerrorintheovertimecalculation—possiblycomputing6×750.00) results from a small arithmetic error in the overtime calculation — possibly computing 6 × 750.00)resultsfromasmallarithmeticerrorintheovertimecalculation—possiblycomputing6×25 = 150instead.ChoiceD(150 instead. Choice D (150instead.ChoiceD(1,035.00) applies the overtime rate of 22.50toall46hours,asiftheentireshiftwereovertime:46×22.50 to all 46 hours, as if the entire shift were overtime: 46 × 22.50toall46hours,asiftheentireshiftwereovertime:46×22.50 = $1,035. Pro tip: Always split overtime problems into two separate calculations — regular hours at base pay, overtime hours at the elevated rate — then add. Never apply one rate to the full total.

Question 14

At 3:00 PM, a flagpole casts a 24-foot shadow and a 6.0-foot-tall man casts a 4.0-foot shadow. What is the height, in feet, of the flagpole?

  1. 16
  2. 24
  3. 36 (correct answer)
  4. 42

Explanation: This is a similar triangles and proportions question. Choice C (36) is correct — at the same time of day, the sun's angle is the same, so the person and flagpole form similar right triangles with their shadows. Set up the proportion: height/shadow = 6/4 = h/24. Cross-multiply: 4h = 144 → h = 36 feet. Choice A (16) results from inverting the proportion: h/24 = 4/6 → h = 24 × (4/6) = 16. Choice B (24) gives the shadow length — the student reports the shadow as the height. Choice D (42) adds the man's height to the pole's shadow: 6 + 36 = 42 or 24 + 18 = 42. Pro tip: In shadow problems, both objects cast shadows at the same angle. Always set up the proportion as height/shadow = height/shadow (same ratio for both). Label clearly: (person's height)/(person's shadow) = (tree's height)/(tree's shadow), then solve for the unknown.

Question 15

A polynomial is given by a(x)=9−5x+x3a(x) = 9 - 5x + x^3a(x)=9−5x+x3. What is the degree of a(x)a(x)a(x)?

  1. 999
  2. 111
  3. 333 (correct answer)
  4. 555

Explanation: The degree of a polynomial is the highest exponent of the variable. In a(x) = 9 - 5x + x³, rewrite in standard form as a(x) = x³ - 5x + 9. The terms have exponents: x³ has exponent 3, -5x has exponent 1, and 9 has exponent 0. The highest exponent is 3, so the degree is 3.

Question 16

A rectangular prism has length 8 units, width 5 units, and height 3 units. What is the volume of the rectangular prism?

  1. 120 cubic units120\text{ cubic units}120 cubic units (correct answer)
  2. 94 cubic units94\text{ cubic units}94 cubic units
  3. 158 cubic units158\text{ cubic units}158 cubic units
  4. 46 cubic units46\text{ cubic units}46 cubic units

Explanation: We need to find the volume of a rectangular prism with length 8 units, width 5 units, and height 3 units. The volume formula for a rectangular prism is V = l × w × h. Substituting the given dimensions: V = 8 × 5 × 3 = 40 × 3 = 120 cubic units.

Question 17

Which of the following is the product ABABAB if A=[0231]A = \begin{bmatrix} 0 & 2 \\ 3 & 1 \end{bmatrix}A=[03​21​] and B=[1−140]B = \begin{bmatrix} 1 & -1 \\ 4 & 0 \end{bmatrix}B=[14​−10​]?

  1. [8034]\begin{bmatrix} 8 & 0 \\ 3 & 4 \end{bmatrix}[83​04​]
  2. [80312]\begin{bmatrix} 8 & 0 \\ 3 & 12 \end{bmatrix}[83​012​]
  3. [807−3]\begin{bmatrix} 8 & 0 \\ 7 & -3 \end{bmatrix}[87​0−3​] (correct answer)
  4. [8073]\begin{bmatrix} 8 & 0 \\ 7 & 3 \end{bmatrix}[87​03​]

Explanation: This question involves matrix multiplication, where entry (i,j) of AB equals the dot product of row i of A with column j of B. For entry (1,1): (0)(1) + (2)(4) = 0 + 8 = 8, for (1,2): (0)(-1) + (2)(0) = 0 + 0 = 0, for (2,1): (3)(1) + (1)(4) = 3 + 4 = 7, and for (2,2): (3)(-1) + (1)(0) = -3 + 0 = -3. The result is [[8,0],[7,-3]].

Question 18

Triangles △ABC\triangle ABC△ABC and △DEF\triangle DEF△DEF are shown. ∠A≅∠D\angle A \cong \angle D∠A≅∠D and ∠B≅∠E\angle B \cong \angle E∠B≅∠E are marked. Also, the side BCBCBC is marked congruent to EFEFEF (one tick), but it is not the included side between the marked angles. Which congruence criterion applies?

  1. ASA
  2. SAS
  3. AAS (correct answer)
  4. SSS

Explanation: The triangles are congruent by AAS since two pairs of corresponding angles are congruent (∠A≅∠D and ∠B≅∠E)(\angle A \cong \angle D \text{ and } \angle B \cong \angle E)(∠A≅∠D and ∠B≅∠E) and a pair of corresponding sides not between the angles is congruent (BC≅EF)(BC \cong EF)(BC≅EF). In AAS, the congruent side is not the included side between the two marked angles. Since BCBCBC is not between ∠A\angle A∠A and ∠B\angle B∠B, and EFEFEF is not between ∠D\angle D∠D and ∠E\angle E∠E, the AAS criterion applies.

Question 19

If v=⟨5,12⟩\mathbf{v} = \langle 5, 12 \ranglev=⟨5,12⟩, what is −v-\mathbf{v}−v?

  1. ⟨0,0⟩\langle 0, 0 \rangle⟨0,0⟩
  2. ⟨5,−12⟩\langle 5, -12 \rangle⟨5,−12⟩
  3. ⟨−5,12⟩\langle -5, 12 \rangle⟨−5,12⟩
  4. ⟨−5,−12⟩\langle -5, -12 \rangle⟨−5,−12⟩ (correct answer)

Explanation: This problem asks for the negative of a vector. The negative of vector ⟨a, b⟩ is ⟨-a, -b⟩. For -v = -⟨5, 12⟩, we negate each component: ⟨-5, -12⟩. The negative vector has the same magnitude but opposite direction.

Question 20

Lines rrr and sss are parallel, cut by transversal ttt. If angle 111 is 75o75^\text{o}75o, what is the measure of angle 444, the alternate exterior angle?

  1. 105°
  2. 75° (correct answer)
  3. 85°
  4. 95°

Explanation: Angles 1 and 4 are alternate exterior angles formed by parallel lines r and s cut by transversal t. When parallel lines are cut by a transversal, alternate exterior angles are equal. Since angle 1 is 75°, angle 4 must also be 75°. Choice A incorrectly uses 105°, which would be the supplementary angle.

Question 21

A county wants to estimate the percentage of households with reliable internet. They have a complete address list and use a random number generator to select 600 addresses to mail a survey. Some households may not respond, creating nonresponse bias. Which sampling method is being used?

  1. Simple random sample by randomly selecting addresses from the full county list (correct answer)
  2. Convenience sample by choosing addresses closest to the county office
  3. Cluster sample by selecting one neighborhood and surveying every house in it
  4. Systematic sample by choosing every address with an even house number

Explanation: This is a simple random sample because the county uses a random number generator to select addresses from the complete population list. Simple random sampling gives every unit in the population an equal probability of being selected through a random process. The county follows this procedure exactly by using random selection from their comprehensive address list. Choice C incorrectly describes cluster sampling, which would involve selecting entire neighborhoods rather than individual addresses.

Question 22

What is −2v-2\mathbf{v}−2v if v=⟨9,−4⟩\mathbf{v}=\langle 9, -4\ranglev=⟨9,−4⟩?

  1. ⟨−18,8⟩\langle -18, 8\rangle⟨−18,8⟩ (correct answer)
  2. ⟨18,8⟩\langle 18, 8\rangle⟨18,8⟩
  3. ⟨−18,−8⟩\langle -18, -8\rangle⟨−18,−8⟩
  4. ⟨−2,8⟩\langle -2, 8\rangle⟨−2,8⟩

Explanation: This problem involves scalar multiplication. For scalar multiplication, k times ⟨a,b⟩ equals ⟨ka,kb⟩. We calculate -2v = -2⟨9,-4⟩ = ⟨-2×9, -2×(-4)⟩ = ⟨-18,8⟩.

Question 23

Events A and B are mutually exclusive. P(A)=0.5P(A) = 0.5P(A)=0.5 and P(B)=0.3P(B) = 0.3P(B)=0.3. What is P(A or B)P(A \text{ or } B)P(A or B)?

  1. 0.15
  2. 0.2
  3. 0.8 (correct answer)
  4. 1

Explanation: This is a probability question testing the addition rule for mutually exclusive events. Choice C (0.80) is correct — for mutually exclusive events (which cannot both occur), P(A or B) = P(A) + P(B) = 0.5 + 0.3 = 0.8. Choice A (0.15) multiplies the probabilities: P(A) × P(B) = 0.5 × 0.3 = 0.15 — this is the formula for P(A and B) when events are INDEPENDENT, not the formula for P(A or B) when events are mutually exclusive. Choice B (0.20) subtracts: 0.5 − 0.3 = 0.2. Choice D (1.00) assumes mutually exclusive events together cover the entire sample space — but two mutually exclusive events can have probabilities that sum to less than 1 (there can be other outcomes). Pro tip: Mutually exclusive means the events cannot happen at the same time — like rolling a 2 and rolling a 5 on the same die. For mutually exclusive events: P(A or B) = P(A) + P(B). For non-mutually-exclusive events: P(A or B) = P(A) + P(B) − P(A and B). The simpler formula here is a gift — just add.

Question 24

Solve the system:

x + y = 8\\ 3x - y = 4 \end{cases}$$ What is the solution $(x, y)$?
  1. (1,7)(1, 7)(1,7)
  2. (4,4)(4, 4)(4,4)
  3. (2,6)(2, 6)(2,6)
  4. (3,5)(3, 5)(3,5) (correct answer)

Explanation: Use elimination: Add the equations x+y=8x + y = 8x+y=8 and 3x−y=43x - y = 43x−y=4. This gives 4x=124x = 124x=12, so x=3x = 3x=3. Substitute x=3x = 3x=3 into x+y=8x + y = 8x+y=8: 3+y=83 + y = 83+y=8, so y=5y = 5y=5. The solution is (3,5)(3, 5)(3,5).

Question 25

Let P(t)=400(3)t2P(t) = 400(3)^{\frac{t}{2}}P(t)=400(3)2t​. What is the value of P(6)P(6)P(6)?

  1. 2,400
  2. 3,600
  3. 10,800 (correct answer)
  4. 32,400

Explanation: This is an exponential models question testing careful evaluation of a fractional exponent. Choice C (10,800) is correct — substitute t = 6: P(6) = 400(3)^(6/2) = 400(3)³ = 400 × 27 = 10,800. Choice A (2,400) treats the exponent as a multiplier: 400 × 6 = 2,400, ignoring the exponential structure entirely. Choice B (3,600) uses exponent 2 instead of 3: 400 × 3² = 400 × 9 = 3,600 — computing 6/2 as 2 rather than 3, or using n − 1 = 2 from sequence thinking. Choice D (32,400) uses exponent 4 instead of 3: 400 × 3⁴ = 400 × 81 = 32,400 — perhaps computing (6/2) + 1 = 4. Pro tip: Always resolve the exponent completely before computing the power. Here, t/2 = 6/2 = 3, so the base 3 is raised to the 3rd power: 3³ = 27. Writing it out as 400 × 3 × 3 × 3 = 400 × 27 avoids confusion about what the exponent is.