1.
Question 1. What is the degree of the function \(f(x)=5x-4x^2-x^3\)?
Solution: The degree of a polynomial is the highest exponent of \(x\). In \(f(x)=5x-4x^2-x^3\), the highest power is \(3\). Answer: B
2.
Question 2. Find the equation of a line passing through the points \((-1,5)\) and \((3,17)\).
Solution: First find the slope: \[ m=\frac{y_2-y_1}{x_2-x_1}=\frac{17-5}{3-(-1)}=\frac{12}{4}=3. \] Use point-slope form with \((-1,5)\): \[ y-5=3(x-(-1)) \] \[ y-5=3x+3 \] \[ y=3x+8 \] Answer: A
3.
Question 3. The amplitude of \(y=4\cos(x+\pi)\) is \(4\).
Solution: For \(y=a\cos(b(x-c))+d\), the amplitude is \(|a|\). Here \(a=4\), so the amplitude is: \[ |4|=4 \] The statement is true. Answer: A
4.
Question 4. Find the inverse of \(f(x)=x^7\).
Solution: Let \(y=x^7\). Switch \(x\) and \(y\): \[ x=y^7 \] Solve for \(y\): \[ y=\sqrt[7]{x}=x^{\frac{1}{7}} \] So the inverse is \(f^{-1}(x)=x^{\frac{1}{7}}\). Answer: C
5.
Question 5. The functions \(f(x)=\frac{1}{x^3}\) and \(g(x)=x^{-\frac{1}{3}}\) are inverses of each other.
Solution: Check the inverse of \(y=\frac{1}{x^3}\). Switch \(x\) and \(y\): \[ x=\frac{1}{y^3} \] \[ y^3=\frac{1}{x} \] \[ y=\sqrt[3]{\frac{1}{x}}=x^{-\frac{1}{3}} \] This matches \(g(x)=x^{-\frac{1}{3}}\), so the functions are inverses. Answer: B
6.
Question 6. The value of \(\log_4(120)\) to two decimal places is \(3.45\).
Solution: Use the change of base formula: \[ \log_4(120)=\frac{\log(120)}{\log(4)} \] \[ \log_4(120)\approx 3.45 \] The statement is true. Answer: A
7.
Question 7. Solving for \(x\) in the equation \(2^{-x}=16\) gives us \(-4\).
Solution: Rewrite \(16\) as a power of \(2\): \[ 2^{-x}=16=2^4 \] Since the bases are the same: \[ -x=4 \] \[ x=-4 \] The statement is true. Answer: A
8.
Question 8. Determine the equation of a line, in slope-intercept form, that passes through the points \((-2,-1)\) and \((-4,4)\).
Solution: First find the slope: \[ m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-(-1)}{-4-(-2)}=\frac{5}{-2}=-\frac{5}{2} \] Use point-slope form with \((-2,-1)\): \[ y-(-1)=-\frac{5}{2}(x-(-2)) \] \[ y+1=-\frac{5}{2}(x+2) \] \[ y+1=-\frac{5}{2}x-5 \] \[ y=-\frac{5}{2}x-6 \] Answer: C
9.
Question 9. Given an equation for a transformed sine function with an amplitude of \(\frac{1}{3}\), a period of \(\frac{3}{2}\), a phase shift of \(\frac{5}{6}\) rad to the right, and a vertical translation of \(4\) units down.
Solution: Use the transformed sine form: \[ y=a\sin(b(x-c))+d \] The amplitude is \(|a|=\frac{1}{3}\), so \(a=\frac{1}{3}\). The period is \(\frac{3}{2}\), so: \[ b=\frac{2\pi}{\text{period}}=\frac{2\pi}{\frac{3}{2}}=\frac{4\pi}{3} \] A phase shift of \(\frac{5}{6}\) to the right gives \(x-\frac{5}{6}\). A vertical translation of \(4\) units down gives \(-4\). Therefore: \[ y=\frac{1}{3}\sin\left(\frac{4\pi}{3}\left(x-\frac{5}{6}\right)\right)-4 \] Answer: A
10.
Question 10. Which of the following is equivalent to the expression \(\log_9(r)+4\log_9(w)+\log_9(x)\)?
Solution: Use the power rule of logarithms: \[ 4\log_9(w)=\log_9(w^4) \] Then use the product rule: \[ \log_9(r)+\log_9(w^4)+\log_9(x)=\log_9(rw^4x) \] Answer: D
11.
Question 11. Evaluate \(\log_8(32768)\).
Solution: Since: \[ 8^5=32768 \] we have: \[ \log_8(32768)=5 \] Answer: C
12.
Question 12. Solve \(\log_2(x)=\log_2(2)+\log_2(5)\).
Solution: Use the product rule of logarithms: \[ \log_2(2)+\log_2(5)=\log_2(10) \] So: \[ \log_2(x)=\log_2(10) \] Therefore: \[ x=10 \] Answer: A
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