SAT Math — Module 2 (harder) practice test

The initial area is 80, and doubling each hour gives a growth factor of 2. Therefore, $A=80(2)^t$.

The equation $|2x+1|=9$ gives $2x+1=9$ or $2x+1=-9$, so $x=4$ or $x=-5$. The positive solution is 4.

The machine loses $14 over 10 months at a constant rate, so its value decreases by $\dfrac{14}{10}=1.4$ dollars per month.

When powers with the same base are multiplied, their exponents are added. Thus, the expression becomes $h^{3+2}j^{-2+6}=h^5j^4$.

The left side of $5x-35=-20$ is $5(x-7)$, so $5(x-7)=-20$ and $x-7=-4$.

Substituting $(4,-2)$ into $2x+ky=14$ gives $8-2k=14$. Solving yields $k=-3$.

The volume is $\dfrac13(9^2)(8)=216$.

Relative to angle $G$, $HJ$ is the opposite leg and $GH$ is the adjacent leg. Therefore, $\tan G=\dfrac{HJ}{GH}=\dfrac{39}{80}$.

The coefficients 8 and 3 are the unit prices, in dollars, of a notebook and a pen, respectively. A notebook costs $8-3=5$ dollars more than a pen.

The $101^\circ$ angle and the unmarked exterior angle adjacent to it on line $q$ are supplementary, so that unmarked angle measures $79^\circ$. That unmarked angle and the angle labeled $(3x+7)^\circ$ are alternate exterior angles, so $3x+7=79$ and $x=24$.

Since $(x-4)^2 \ge 0$ and the leading coefficient is positive, the minimum occurs at $x = 4$ and equals 11.

The line through $(0,3)$ and $(10,23)$ is $y=2x+3$, which predicts 17 at $x=7$; the actual value is 19, which is 2 greater.

Subtracting $2\pi$ gives $17\pi/6-12\pi/6=5\pi/6$. This angle lies between $\pi/2$ and $\pi$, so its terminal side lies in quadrant II.

$1.06\times1.04=1.1024$, so the revenue increased by $10.24\%$.

There are 14 values, so the median is the mean of the 7th and 8th values, both of which are 12.

The net rate is $14-5=9$ gallons per minute. In 9 minutes the amount of water in the tank increases by $9\times9=81$ gallons.

For points A through D, the values of $7x-9y$ are 1271, 1264, 1270, and 1273, respectively. Exactly three values exceed $k$ when $1264\le k<1270$, so the greatest possible integer value of $k$ is 1269.

From the table, $f(0)=-12$ and $f(4)=4$, so $f$ has slope 4 and equation $f(x)=4x-12$. Setting $f(x)=0$ gives $x=3$.

The right side is $6(3x-a)=18x-6a$. For $18x-7=18x-6a$ to have infinitely many solutions, the constant terms must match: $-7=-6a$, so $a=\dfrac{7}{6}$.

Expanding gives $kx^2 + 30x + 18 = 0$, with discriminant $30^2 - 4(k)(18) = 900 - 72k$. No real solutions requires $900 - 72k < 0$, so $k > 12.5$. The least integer satisfying this is 13.

Factor the second denominator as $ut(u-2)$, so the second term simplifies to $\dfrac{1}{u}$. Then $\dfrac{t+5}{u-2}+\dfrac{1}{u} = \dfrac{u(t+5)+(u-2)}{u(u-2)} = \dfrac{tu+6u-2}{u^2-2u}$.

The $y$-intercept of $f$ is $f(0) = a + b$. In $g$, substituting $x=0$ gives $g(0) = a(1-1) + n = n$; since $g \equiv f$ requires $n = a+b$, the constant $n$ equals the $y$-intercept directly with no further computation needed, because the $(3^x-1)$ factor vanishes at $x=0$. In $h$, substituting $x=0$ gives $h(0) = 3a\cdot 3^{-1} + p = a + p$, which still requires adding two nonzero terms ($a$ and $p$) to get the $y$-intercept — $p$ alone is not the $y$-intercept. So only $g$ displays it directly as $n$ or $p$.