SAT Math Full Practice Test, Advanced Level

The answer is (B), $3$ watt-hours per minute.

1. Solve the input rule for $t$. Multiplying by $3$ and adding $4$ gives $t = 3u + 4$, so $u = 2$ and $u = 6$ correspond to $t = 10$ and $t = 22$ minutes. Both satisfy the condition of at least $4$ minutes.

2. The energy changes by $65 - 29 = 36$ watt-hours while the time changes by $22 - 10 = 12$ minutes, so the rate is

   $$\frac{36}{12} = 3 \text{ watt-hours per minute.}$$

3. Dividing by $6 - 2$ instead gives $9$ watt-hours per unit of $u$. Each unit of $u$ is $3$ minutes, so that $9$ must be divided by $3$. The added $4$ shifts the start of the time scale and does not make each interval longer.

4. As a check, the rule $E = 3t - 1$ gives $3 \cdot 10 - 1 = 29$ and $3 \cdot 22 - 1 = 65$. The model is used only on the stated domain, so its intercept is not read as the energy used at time zero.

Choice (A) is the rate with respect to $u$, $\frac{\Delta E}{\Delta u}$, not with respect to $t$. Choice (C) reports the time difference as a rate. Choice (D) multiplies the rate per unit of $u$ by $3$ instead of dividing.

The answer is (B), $4 < k < 6$.

1. Subtract $2x$ and $3$ from both sides. This collects $x$ without dividing by a quantity that might be zero and gives $(k - 4)x = k - 2$.

2. At $k = 4$ the equation says $0 \cdot x = 2$, which no real $x$ satisfies. For every other $k$,

   $$x = \frac{k - 2}{k - 4} = 1 + \frac{2}{k - 4}.$$

3. For $x > 2$ the fraction $\frac{2}{k - 4}$ must exceed $1$. Its numerator is positive, so its denominator must be positive, which means $k > 4$. Multiplying by the positive $k - 4$ keeps the direction of the inequality, so $2 > k - 4$, which gives $k < 6$.

4. Together the conditions give $4 < k < 6$. At $k = 6$ the solution is $x = 2$, which fails the strict condition, and $k = 4$ gives no solution, so neither endpoint belongs.

5. As a check inside the interval, $k = 5$ gives $x = 3$. The left side is $(5 - 2) \cdot 3 + 3 = 12$ and the right side is $2 \cdot 3 + 5 + 1 = 12$.

Choice (A) fails because for $k < 4$ the fraction $\frac{2}{k - 4}$ is negative, which gives $x < 1$. Choice (C) fails because for $k > 6$ that positive fraction is less than $1$, which gives $x < 2$. Choice (D) leaves out the case with no solution but ignores the condition $x > 2$.

The answer is (C), $-4$.

1. Complete the square to show the lowest point.

   $$p(x) = 3(x^2 - 6x) + 31 = 3\left[(x - 3)^2 - 9\right] + 31 = 3(x - 3)^2 + 4.$$

2. The equation $p(x) + m = 0$ becomes $3(x - 3)^2 = -4 - m$.

3. The left side is never negative and is zero only at $x = 3$. A positive right side gives two solutions, zero gives one, and a negative right side gives none.

4. Exactly one solution needs $-4 - m = 0$, so $m = -4$. The requested value is the shift $m$, and $x = 3$ is the root that this shift produces.

5. As a check, $3x^2 - 18x + 31 - 4 = 3x^2 - 18x + 27 = 3(x - 3)^2$, which is zero only at $x = 3$.

Choice (A) makes the constant term zero, and $3x^2 - 18x = 0$ has the two roots $0$ and $6$. Choice (B) uses the minimum value instead of its negative, which raises the minimum from $4$ to $8$ and leaves no real root. Choice (D) uses the constant $9$ from inside the brackets as the shift.

The answer is $45$.

1. Look for a combination of the whole equations that produces the target, instead of finding $u$ and $v$ first. Twice the first left side is $4u + 6v$, and adding $5u - 2v$ gives exactly $9u + 4v$.

2. Doubling the first equation doubles its right side too, from $17$ to $34$. Adding the two equations gives

   $$9u + 4v = 34 + 11 = 45.$$

3. As a separate check, multiply the first equation by $2$ and the second by $3$ and add them, which gives $19u = 67$. So $u = \frac{67}{19}$ and $v = \frac{63}{19}$. These values satisfy both original equations, and $9u + 4v = \frac{855}{19} = 45$.

4. A common slip is to double only the left side. Scaling the whole equation is what makes the shorter method valid.

The answer is (A), $90.

1. Let $P$ be the original price. The increase leaves $120\%$ of $P$. The discount is taken from the increased price and leaves $75\%$ of it, so

   $$0.75 \cdot 1.2P = 0.9P = 81.$$

2. Dividing by $0.9$ gives $P = 90$. Multiplying $81$ by $0.9$ would apply another reduction instead of undoing one.

3. As a check, $20\%$ of $90 is $18, which gives $108. Then $25\%$ of $108 is $27, which leaves $81. The two changes differ in dollars because their bases differ.

4. Subtracting $25\%$ from $20\%$ would suggest a $5\%$ decrease on one common base. The actual net loss is $9 from $90, which is $10\%$. The amount $108 is a real intermediate price, but the question asks for the original price.

Choice (B) recovers the price before the discount but does not undo the increase. Choice (C) adds $25\%$ of the final price to undo the discount, which uses the wrong base. Choice (D) treats the changes as a net $5\%$ decrease, solves $0.95P = 81$ and rounds the incorrect result.

The answer is (B), $14$.

1. Triangles $AEB$ and $CED$ have equal vertical angles at $E$. Because $AB \parallel CD$, the angle between $AB$ and $AE$ equals the angle between $CD$ and $CE$. Two pairs of equal angles make the triangles similar, with $A$ matching $C$, $E$ matching $E$ and $B$ matching $D$.

2. The side $AB$ matches $CD$ and $AE$ matches $CE$, so

   $$\frac{6}{15} = \frac{4}{CE}.$$

   This gives $6 \cdot CE = 60$, so $CE = 10$.

3. The diagonal $AC$ is made of $AE$ and $EC$, so $AC = 4 + 10 = 14$. The length $10$ alone stops at the intersection.

4. As a check, $\frac{6}{15}$ and $\frac{4}{10}$ both reduce to $\frac{2}{5}$. The ratio rests on the given parallel sides and the matching angles, not on how the sketch looks.

Choice (A) finds $CE$ but leaves out $AE$. Choice (C) adds $AE$ to the base $CD$, which is not part of the diagonal. Choice (D) adds the two parallel bases.

The answer is (C), $1{,}280$.

1. Divide the two readings. Since $a$ and $b$ are positive, the common factors cancel and

   $$\frac{N(7)}{N(1)} = \frac{ab^7}{ab} = b^6 = \frac{640}{80} = 8.$$

2. The step from hour $7$ to hour $9$ takes two hours, with multiplier $b^2$. Three two-hour intervals make six hours, so $(b^2)^3 = 8$ and $b^2 = 2$. The quantity doubles every two hours.

3. So $N(9) = 640 \cdot 2 = 1{,}280$. Applying the six-hour multiplier $8$ again would move six hours ahead, not two.

4. As a check, starting from $80$ at hour $1$ and doubling every two hours gives $160$, $320$, $640$ and $1{,}280$ at hours $3$, $5$, $7$ and $9$. Equal time intervals use equal multipliers, not equal added amounts.

Choice (A) adds the starting amount $80$ to $640$ instead of using a multiplier. Choice (B) uses an unsupported multiplier of $1.5$. Choice (D) applies the six-hour multiplier $8$ to a two-hour change.

The answer is $4$.

1. Factor the numerator as $(x - 3)(x + 3)$. On the allowed domain $x \ne 3$ the factor $x - 3$ cancels, which leaves $x + 3 = k(x - 1)$. The restriction $x \ne 3$ stays with the new equation.

2. Rearranging gives $(k - 1)x = k + 3$. At $k = 1$ this says $0 \cdot x = 4$, which is impossible, so $1$ is in $S$.

3. For $k \ne 1$ the only candidate is

   $$x = \frac{k + 3}{k - 1}.$$

   It is unusable when it equals $3$. Multiplying by the nonzero $k - 1$ gives $k + 3 = 3k - 3$, so $k = 3$. At $k = 3$ the only candidate is the excluded $x = 3$, so $3$ is in $S$.

4. For every other $k$ the candidate differs from $3$, so the equation has one allowed solution. Thus $S = \{1, 3\}$, and the sum of its values is $4$.

5. As a check outside $S$, $k = 0$ gives $x = -3$. In the original equation the left side is $\frac{0}{-6} = 0$ and the right side is $0$ too.

The answer is (B), $7$.

1. With $n$ packs the conditions are $18 + 7n \le 100$ and $n \ge 5$.
2. Subtracting $18$ gives $7n \le 82$, and dividing by the positive $7$ gives $n \le \frac{82}{7} = 11\frac{5}{7}$. Whole packs make $11$ the largest count.
3. The allowed counts are $5$, $6$, $7$, $8$, $9$, $10$ and $11$, which is $7$ different counts. The difference $11 - 5 = 6$ counts the gaps between them, and both endpoints are allowed.
4. As a check, $5$ packs cost $53 and $11$ packs cost $95, while $12$ packs cost $102 and exceed the budget.

Choice (A) counts the six gaps instead of the seven allowed values. Choice (C) is the largest allowed count, not the number of possible counts. Choice (D) rounds the upper bound up and also confuses a value with a count.

The answer is (D), the reported $95\%$ confidence interval is $1{,}320$ to $1{,}720$ registered members.

1. The interval for the preference is $62\%$ plus or minus $5$ percentage points, from $57\%$ to $67\%$. Five percentage points is a direct difference on the percent scale, not $5\%$ of $62$.
2. The question asks about members who do not prefer the workshops. Subtracting from $100\%$ reverses which endpoint is lower and gives $33\%$ to $43\%$.
3. Multiplying by the $4{,}000$ registered members gives $0.33 \cdot 4{,}000 = 1{,}320$ and $0.43 \cdot 4{,}000 = 1{,}720$.
4. The middle estimate is $38\%$ of $4{,}000$, or $1{,}520$ members, but a sample does not establish an exact population count. The confidence level describes how this method of making intervals performs over repeated random samples, and the interval is about registered members, not all city residents.

Choice (A) treats the point estimate for the complement as a known exact count. Choice (B) counts the members who prefer Saturday workshops, not the requested group. Choice (C) extends the inference beyond the registered members the sample was drawn from.

The answer is (A), $-1$.

1. Parallel lines have proportional $x$ and $y$ coefficients. To avoid dividing by $k - 3$, which might be zero, compare cross products, $4(k - 1) = 2(k - 3)$.
2. Expanding gives $4k - 4 = 2k - 6$, so $2k = -2$ and $k = -1$. This is the only value that makes the coefficients proportional.
3. At $k = -1$ the first line is $-2x - 4y = 2$, or $2x + 4y = -2$. The second line is $2x + 4y = 5$, so the lines share no point and are distinct.
4. In slope-intercept form the lines are $y = -\frac{1}{2}x - \frac{1}{2}$ and $y = -\frac{1}{2}x + \frac{5}{4}$, with the same slope and different intercepts.
5. The coefficients $k - 1$ and $k - 3$ are never both zero, so the first equation is always a line. At $k = 3$ it is vertical while the second line is not, so that case fails anyway.

Choice (B) makes the first line horizontal, which does not match the second line. Choice (C) makes the first line the vertical line $x = 1$, which cannot be parallel to the second line. Choice (D) gives the first line slope $-2$ instead of $-\frac{1}{2}$.

The answer is $-3$.

1. Because $x - 2$ is a factor, $P(2) = 0$. That gives $8 + 4a + 2b - 12 = 0$, so $4a + 2b = 4$, or $2a + b = 2$.
2. The condition $P(-1) = 0$ gives $-1 + a - b - 12 = 0$, so $a - b = 13$.
3. Adding the two equations gives $3a = 15$, so $a = 5$, and then $b = 2 - 10 = -8$. The requested sum is $a + b = -3$.
4. As a check, $(x - 2)(x + 1)(x + 6) = x^3 + 5x^2 - 8x - 12$, which has the original form, both required zeros and the coefficients found.

The answer is (D), $2$ millimeters.

1. A sum is a mean times a count. Group R starts with a sum of $20 \cdot 14 = 280$ millimeters and group S with $30 \cdot 22 = 660$ millimeters.
2. The moved measurements total $10 \cdot 26 = 260$ millimeters. Group R now has $30$ measurements with sum $280 + 260 = 540$, so its mean is $\frac{540}{30} = 18$ millimeters.
3. Group S has $20$ measurements left with sum $660 - 260 = 400$, so its mean is $\frac{400}{20} = 20$ millimeters.
4. The mean of S is greater by $20 - 18 = 2$ millimeters.
5. As a check, the total is still $540 + 400 = 940$ millimeters over $30 + 20 = 50$ measurements. The old gap was $8$ and it shrank by $6$, but the question asks for the new gap.

Choice (A) uses the new mean of R but the old mean of S. Choice (B) is the decrease in the gap, $8 - 2$, not the new gap. Choice (C) keeps the original gap $22 - 14$ despite the move.

The answer is (A), $5$.

1. Complete both squares by adding $16$ and $9$ to both sides, which gives $(x - 4)^2 + (y + 3)^2 = 25$. The center is $C = (4, -3)$ and the radius is $5$.
2. From the center to $P$ the horizontal change is $6$ and the vertical change is $8$, so $CP = \sqrt{6^2 + 8^2} = 10$. The point $P$ is outside the circle.
3. For any point $Q$ on the circle, $CP \le CQ + QP$, so $QP \ge 10 - 5 = 5$. The point of the circle on the segment from $C$ to $P$ reaches this distance, so the shortest distance is $5$.
4. As a check, that point is $(7, 1)$, halfway from the center to $P$. It satisfies $49 + 1 - 56 + 6 = 0$, and its distance to $P$ is $\sqrt{3^2 + 4^2} = 5$.

Choice (B) is the distance to the center, not to the circle. Choice (C) is the distance to the farthest point of the circle. Choice (D) is the radius squared from the completed square equation.

The answer is (C), $3$.

1. Both equations are lines because their $y$ coefficients are not zero. Infinitely many common solutions means they are the same line, so one whole equation, constant included, is a nonzero multiple of the other.
2. The $y$ coefficients fix the multiplier at $\frac{6}{4} = \frac{3}{2}$. Multiplying the whole second equation by $\frac{3}{2}$ gives $3x + 6y = 9$.
3. Matching the $x$ coefficient requires $k = 3$. Then the first right side is $3 + 6 = 9$, which also matches, so the lines coincide.
4. Equal coefficient ratios alone would allow parallel lines with different constants, which is why the constant has to be checked as well.
5. At $k = 3$ both equations reduce to $y = \frac{3}{2} - \frac{1}{2}x$, so every real $x$ gives a solution, for example $\left(0, \frac{3}{2}\right)$ and $(3, 0)$.

Choice (A) misses the case $k = 3$. Choice (B) matches the $x$ coefficient of the second equation without using the multiplier fixed by the $y$ coefficients. Choice (D) confuses the $y$ coefficient of the first equation with the needed $x$ coefficient.

The answer is $8$.

1. A real square root is never negative, so the right side needs $x - 3 \ge 0$, or $x \ge 3$. This also makes $2x + 9$ positive.
2. Squaring both sides gives $2x + 9 = x^2 - 6x + 9$, so $x^2 - 8x = 0$ and $x(x - 8) = 0$. The candidates are $0$ and $8$. Factoring instead of dividing by $x$ keeps the candidate $0$ available for checking.
3. The candidate $0$ is below $3$. Directly, $\sqrt{9} = 3$ while $0 - 3 = -3$. Squaring made the two sides look equal because both squares are $9$.
4. At $x = 8$ the left side is $\sqrt{25} = 5$ and the right side is $8 - 3 = 5$, so $8$ is the only real solution.

The answer is (C), $9$.

1. The first rule reaches at most $12 + 3 \cdot 4 = 24$ dollars on its interval, so it cannot give $34.
2. The second rule gives $24 + 2(w - 4) = 34$, so $2(w - 4) = 10$ and $w - 4 = 5$. Those $5$ kilograms are the weight above $4$, so $w = 9$.
3. The weight $9$ is greater than $4$ and at most $10$, so it lies in the interval of the rule used. As a check, $24 + 2(9 - 4) = 34$.
4. Solving a branch equation alone can give a number outside that branch, so both the interval and the charge need checking.

Choice (A) reports only the kilograms above $4$. Choice (B) uses the difference $34 - 24$ with a wrong threshold or rate. Choice (D) is beyond the allowed maximum of $10$ and does not give $34.

The answer is (B), $1$.

1. Subtract $8$ from both sides and add $9$ to complete the square, which gives $(x - 3)^2 = 3$. So $x = 3 - \sqrt{3}$ or $x = 3 + \sqrt{3}$.
2. Since $1 < \sqrt{3} < 2$, the value $3 - \sqrt{3}$ lies between $1$ and $2$ and is allowed, while $3 + \sqrt{3}$ lies between $4$ and $5$ and is not.
3. On a graph, $y = (x - 3)^2 - 1$ meets the line $y = 2$ twice, but only one of the intersections lies in the strip $1 \le x \le 4$.
4. At the endpoints $x = 1$ and $x = 4$ the left side equals $3$ and $0$, so neither endpoint adds a solution.

Choice (A) discards both irrational candidates although the smaller one is allowed. Choice (C) counts both roots without the interval. Choice (D) is more roots than this quadratic equation can have.

The answer is (C), $9$ units.

1. Draw the perpendicular from $O$ to the chord and call its foot $M$. Because $OA = OB$, it splits the chord in half, so $AM = 15$.
2. In the right triangle $OMA$ the Pythagorean theorem gives $OM^2 + 15^2 = 17^2$, so $OM^2 = 289 - 225 = 64$ and $OM = 8$.
3. The perpendicular from the center to the tangent has the length of the radius, $17$. The chord lies between the center and this tangent, so the gap is $17 - 8 = 9$ units. Adding $8$ would measure to the tangent on the other side of the center, which the wording excludes.
4. As a check with coordinates, put $O$ at $(0, 0)$ and the chord endpoints at $(-15, 8)$ and $(15, 8)$, both $17$ units from $O$. The tangent is the line $y = 17$ and the chord lies on $y = 8$, nine units apart.

Choice (A) is the distance from the center to the chord. Choice (B) is half the length of the chord. Choice (D) is $17 + 8$, the gap to the tangent on the other side of the center.

The answer is (C), a causal comparison for the study participants, with limited generalization to all city students.

1. Recruitment and assignment are two different stages. The $120$ participants were volunteers, and random assignment split them between the programs after they had entered the study.
2. With the conditions the same apart from the program, random assignment helps prevent a systematic difference in who receives each program. That makes a causal comparison between the programs possible for these participants. Chance differences can still occur.
3. A higher group mean summarizes the group. It does not show that every participant in A did better, and without information about uncertainty it does not by itself establish a conclusive difference.
4. Volunteers might differ from other city students in ways that matter for tutoring, and randomly assigning them does not make them a random sample of the city. Generalizing to all city students is limited by how the participants were recruited.

Choice (A) generalizes beyond a volunteer sample that was not random. Choice (B) confuses random assignment to programs with random sampling from the population. Choice (D) turns a comparison of means into a guarantee for every individual.

The answer is $-17$.

1. Replace the whole $C$ in the sensor rule by the conversion, keeping the parentheses so the shift stays attached.

   $$E = \frac{9}{5} \cdot \frac{5}{9}(F - 32) + 14 = F - 32 + 14 = F - 18.$$

2. Comparing with $E = pF + q$ gives $p = 1$ and $q = -18$, so $p + q = -17$. The question asks for the sum, not the intercept alone.

3. As a check, $50$ degrees Fahrenheit is $\frac{5}{9} \cdot 18 = 10$ degrees Celsius. The original rule gives $\frac{9}{5} \cdot 10 + 14 = 32$, and the new rule gives $50 - 18 = 32$.

4. Keeping the old constant $14$ after changing the input would miss the subtraction of $32$, and the check would then fail.

The answer is (D), $8$.

1. For $x > 1$ the right side $x + 1$ is positive. The inside expression $2x - 5$ equals either $x + 1$ or $-(x + 1)$.
2. The first case gives $x = 6$. The second case gives $2x - 5 = -x - 1$, so $3x = 4$ and $x = \frac{4}{3}$.
3. Both values exceed $1$, and their product is $6 \cdot \frac{4}{3} = 8$. Their sum, $\frac{22}{3}$, answers a different question.
4. As a check, at $x = 6$ both sides equal $7$. At $x = \frac{4}{3}$ the inside is $\frac{8}{3} - \frac{15}{3} = -\frac{7}{3}$, whose absolute value $\frac{7}{3}$ equals $\frac{4}{3} + 1$.
5. Solving only the positive case misses the smaller root, and writing $-x + 1$ for $-(x + 1)$ changes the equation.

Choice (A) is only the larger solution. Choice (B) is only the smaller solution. Choice (C) adds the solutions instead of multiplying them.

The answer is (C), $192$.

1. A quadratic with leading coefficient $2$ and roots $r$ and $3r$ can be written

   $$2(x - r)(x - 3r) = 2x^2 - 8rx + 6r^2.$$

2. Matching the constant terms gives $6r^2 = 18$, so $r^2 = 3$.

3. Matching the $x$ coefficients gives $-b = -8r$, so $b = 8r$ and $b^2 = 64r^2 = 64 \cdot 3 = 192$. No decimal approximation is needed.

4. Leaving out the leading $2$ in the factored form would change both the linear and the constant coefficients.

5. As a check with $b = 8\sqrt{3}$, at $x = \sqrt{3}$ the terms are $6 - 24 + 18 = 0$, and at $x = 3\sqrt{3}$ they are $54 - 72 + 18 = 0$.

Choice (A) uses the square of the sum of the roots without the factor from the leading coefficient. Choice (B) accounts for the leading coefficient once instead of squaring it in $b^2$. Choice (D) brings in an extra factor of $2$ after finding the correct square.

The answer is (C), $5.2$.

1. Both inputs lie in the interval of the model. For A the prediction is $2.4 \cdot 5 + 7 = 19$, so its residual is $22 - 19 = 3$.
2. For B the prediction is $2.4 \cdot 8 + 7 = 26.2$, so its residual is $24 - 26.2 = -2.2$. The negative sign records that B lies below the line.
3. The residual for A is greater by $3 - (-2.2) = 5.2$.
4. Using the sizes alone would give $3 - 2.2 = 0.8$ and lose which side of the line B is on. As a check, $19 + 3 = 22$ and $26.2 + (-2.2) = 24$ return the observed values.

Choice (A) subtracts the sizes of the residuals and loses the sign for B. Choice (B) is only the residual for B. Choice (D) is the difference between the observed $y$ values, not between the residuals.

The answer is $5$.

1. With $s$ small kits and $l$ large kits, the conditions are $3s + 5l = 61$ and $s \ge 2l$.
2. Each large kit needs at least two small kits, and such a group uses $5 + 3 + 3 = 11$ meters. Replacing $s$ by its lower bound gives $61 = 3s + 5l \ge 11l$, so $l \le \frac{61}{11}$, which lies between $5$ and $6$. Since $l$ is a whole number, $l \le 5$.
3. Six large kits would need at least $12$ small kits and at least $66$ meters, more cord than there is.
4. Five large kits use $25$ meters, which leaves $36$ meters for $12$ small kits. Since $12 \ge 2 \cdot 5$, this arrangement meets the count rule and uses all the cord.
5. More than five large kits is impossible and five is achievable, so the greatest number is $5$.

The answer is (A), $46$.

1. The perpendicular legs are a base and a height, so the triangle’s area is $\frac{1}{2} \cdot 12 \cdot 20 = 120$.
2. The rectangle gives $w(w + 7) = 120$, so $w^2 + 7w - 120 = 0$ and $(w + 15)(w - 8) = 0$. The condition $w > 0$ selects $w = 8$.
3. The length is $8 + 7 = 15$, and the perimeter is $2(8 + 15) = 46$. The sum $23$ covers only half of the boundary.
4. As a check, $8 \cdot 15 = 120$ and $15 - 8 = 7$. Equal areas do not force equal perimeters, so the perimeter has to come from the rectangle’s own sides.

Choice (B) is half the perimeter. Choice (C) treats all four sides as the width, $4 \cdot 8$. Choice (D) uses the size of the rejected root $-15$ as a width with length $22$, and those dimensions fail the area condition.

The answer is (B), $80\left(\frac{2}{3}\right)^{t/2}$.

1. Since $16 = 2^4$ and $81 = 3^4$, the base is $\frac{16}{81} = \left(\frac{2}{3}\right)^4$. This rewrites the same positive base exactly.

2. For a positive base, raising a power to another power multiplies the exponents, so

   $$M(t) = 80\left(\left(\frac{2}{3}\right)^4\right)^{t/8} = 80\left(\frac{2}{3}\right)^{4t/8} = 80\left(\frac{2}{3}\right)^{t/2}.$$

3. The identity holds for every real $t$ because the base stays positive.

4. As a check, at $t = 8$ both forms give the multiplier $\frac{16}{81}$, and at $t = 0$ both give $80$. Matching a few inputs supports the algebra, but the exact power identity is what proves the equivalence.

Choice (A) uses the exponent $2t$ instead of multiplying $\frac{t}{8}$ by $4$. Choice (C) changes the base without changing the exponent. Choice (D) multiplies the exponent by $4$ but drops the divisor $8$.

The answer is (A), $-7$.

1. Both inputs $8$ and $3$ lie in the domain. The slope is $\frac{79 - 31}{10 - 2} = \frac{48}{8} = 6$, so $g(x) = 31 + 6(x - 2)$.
2. Then $g(8) = 31 + 6 \cdot 6 = 67$ and $g(3) = 31 + 6 = 37$.
3. The expression is $67 - 2 \cdot 37 = 67 - 74 = -7$. The factor $2$ multiplies the output $g(3)$ and does not change the input.
4. As a check, the rule $g(x) = 6x + 19$ returns $31$ at $x = 2$ and $79$ at $x = 10$.