Solutions with detailed explanations to Compass math test practice questions in sample 11.
What is the value of \(2(x^{3} - y^{2}) + 2(x^{3} + y^{2})\) for \(x = 2\) and \(y = -4\)?
Substitute \(x\) and \(y\) by their values and simplify:
\[ \begin{aligned} &2(2^{3} - (-4)^{2}) + 2(2^{3} + (-4)^{2}) \\ &= 2(8 - 16) + 2(8 + 16) \\ &= -16 + 48 = 32 \end{aligned} \]If \(x = 3\), what is the value of \(\frac{6}{x} + 4x - 20\)?
Substitute \(x\) by its value and simplify:
\[ \begin{aligned} \frac{6}{x} + 4x - 20 &= \frac{6}{3} + 4(3) - 20 \\ &= 2 + 12 - 20 = -6 \end{aligned} \]The surface area \(A\) of a cylindrical container is given by the formula:
\[ A = 2\pi r^{2} + 2\pi r h \]where \(r\) is the radius, \(h\) is the height, and \(\pi \approx 3.14\). Which value is closest to the area of a container with radius \(10\) cm and height \(5\) cm?
A) 628
B) 1256
C) 314
D) 3140
E) 942
Substitute \(r = 10\), \(h = 5\), and \(\pi = 3.14\):
\[ \begin{aligned} A &= 2(3.14)(10)^{2} + 2(3.14)(10)(5) \\ &= 628 + 314 = 942 \end{aligned} \]If \(x = -2\) and \(y = 3\), what is the value of \(-x^{-y}\)?
Substitute \(x\) and \(y\):
\[ \begin{aligned} -x^{-y} &= -(-2)^{-3} \\ &= -\frac{1}{(-2)^{3}} \\ &= -\frac{1}{-8} = \frac{1}{8} \end{aligned} \]If \(y = -3\), what is the value of \(\frac{y^{3} - 3}{-y + 3}\)?
Substitute \(y = -3\):
\[ \begin{aligned} \frac{y^{3} - 3}{-y + 3} &= \frac{(-3)^{3} - 3}{-(-3) + 3} \\ &= \frac{-27 - 3}{3 + 3} \\ &= \frac{-30}{6} = -5 \end{aligned} \]What is the value of \(3x^{3} - 2x^{2} + 2\) for \(x = -2\)?
Substitute \(x = -2\):
\[ \begin{aligned} 3(-2)^{3} - 2(-2)^{2} + 2 &= 3(-8) - 2(4) + 2 \\ &= -24 - 8 + 2 = -30 \end{aligned} \]If \(x = 2.7\), what is the value of \(x + 0.75(x - 1.5) + 1.2\)?
Substitute \(x = 2.7\):
\[ \begin{aligned} &2.7 + 0.75(2.7 - 1.5) + 1.2 \\ &= 2.7 + 0.75(1.2) + 1.2 \\ &= 2.7 + 0.9 + 1.2 = 4.8 \end{aligned} \]If \(x = -0.07\) and \(y = 1.2\), what is the value of \(xy - 0.2(y - x)\)?
A) 0.338
B) -0.338
C) 0.3
D) -0.3
E) 0.338
Substitute \(x = -0.07\) and \(y = 1.2\):
\[ \begin{aligned} xy - 0.2(y - x) &= (-0.07)(1.2) - 0.2(1.2 - (-0.07)) \\ &= -0.084 - 0.2(1.27) \\ &= -0.084 - 0.254 = -0.338 \end{aligned} \]Mean arterial pressure (MAP) can be approximated by:
\[ \text{MAP} = DP + \frac{1}{3}(SP - DP) \]Find MAP (rounded to the nearest unit) for \(SP = 101\) mmHg and \(DP = 72\) mmHg.
A) 85
B) 100
C) 90
D) 80
E) 82
Substitute \(SP = 101\) and \(DP = 72\):
\[ \begin{aligned} \text{MAP} &= 72 + \frac{1}{3}(101 - 72) \\ &= 72 + \frac{1}{3}(29) \\ &= 72 + 9.666\ldots \approx 82 \end{aligned} \]What is the value of \(\frac{3}{x - 4y} + \frac{2}{0.5x + 8y}\) for \(x = -2\) and \(y = -0.25\)?
A) \(\frac{11}{3}\)
B) \(\frac{3}{11}\)
C) \(-\frac{3}{11}\)
D) \(-\frac{11}{3}\)
E) 3
Substitute \(x = -2\) and \(y = -0.25\):
\[ \begin{aligned} &\frac{3}{-2 - 4(-0.25)} + \frac{2}{0.5(-2) + 8(-0.25)} \\ &= \frac{3}{-2 + 1} + \frac{2}{-1 - 2} \\ &= \frac{3}{-1} + \frac{2}{-3} \\ &= -3 - \frac{2}{3} = -\frac{11}{3} \end{aligned} \]