1.
Question 1. The amplitude of \(y=4\cos(x+\pi)\) is \(4\).
Solution: For a cosine function in the form \(y=a\cos(bx+c)+d\), the amplitude is \(|a|\). Here, \(a=4\), so the amplitude is \(|4|=4\). Answer: True
2.
Question 2. What is the degree of the function \(f(x)=5x-4x^2-x^3\)?
Solution: The degree of a polynomial is the highest power of \(x\). In \(f(x)=5x-4x^2-x^3\), the highest exponent is \(3\). Therefore, the degree is \(3\). Answer: 3
3.
Question 3. The period of \(y=-3\cos(2x+3\pi)\) is \(\pi\).
Solution: For \(y=a\cos(bx+c)+d\), the period is \(\frac{2\pi}{|b|}\). Here, \(b=2\). \[ \text{Period}=\frac{2\pi}{2}=\pi \] So the statement is true. Answer: True
4.
Question 4. \(\frac{x-6}{3}\) is the inverse of which of the following functions?
Solution: Let the inverse function be \[ y=\frac{x-6}{3}. \] Switch \(x\) and \(y\): \[ x=\frac{y-6}{3} \] Solve for \(y\): \[ 3x=y-6 \] \[ y=3x+6 \] Therefore, the original function is \(y=3x+6\). Answer: \(3x+6\)
5.
Question 5. Find the domain of the inverse of \(f(x)=\sqrt{x-2}\).
Solution: The domain of the inverse function is the range of the original function. For \[ f(x)=\sqrt{x-2}, \] a square root output is always non-negative. So the range of the original function is \[ y\ge 0. \] Therefore, the domain of the inverse is \[ x\ge 0. \] Answer: \(x\ge 0\)
6.
Question 6. \(2^{-3}=?\)
Solution: Use the negative exponent rule: \[ a^{-n}=\frac{1}{a^n} \] So \[ 2^{-3}=\frac{1}{2^3}=\frac{1}{8}. \] Answer: \(\frac{1}{8}\)
7.
Question 7. \(\log_3 27=3\).
Solution: Rewrite \(27\) as a power of \(3\): \[ 27=3^3 \] Then \[ \log_3 27=\log_3(3^3)=3. \] So the statement is true. Answer: True
8.
Question 8. Determine the equation of a line, in slope-intercept form, that passes through points \((3,9)\) and \((6,4)\).
Solution: First find the slope: \[ m=\frac{y_2-y_1}{x_2-x_1} =\frac{4-9}{6-3} =-\frac{5}{3} \] Use point-slope form with \((3,9)\): \[ y-9=-\frac{5}{3}(x-3) \] Expand: \[ y-9=-\frac{5}{3}x+5 \] Add \(9\) to both sides: \[ y=-\frac{5}{3}x+14 \] Answer: \(y=-\frac{5}{3}x+14\)
9.
Question 9. Given an equation for a transformed sine function with an amplitude of \(\frac{4}{7}\), a period of \(3\), a phase shift of \(\frac{7}{2}\) rad to the right, and a vertical translation of \(5\) units down.
Solution: Use the transformed sine form \[ y=a\sin(b(x-c))+d. \] The amplitude is \(|a|=\frac{4}{7}\), so \(a=\frac{4}{7}\). The period is \[ \frac{2\pi}{|b|}=3. \] So \[ b=\frac{2\pi}{3}. \] A phase shift of \(\frac{7}{2}\) to the right means \(x-\frac{7}{2}\). A vertical translation of \(5\) units down means \(d=-5\). Therefore, \[ y=\frac{4}{7}\sin\left(\frac{2\pi}{3}\left(x-\frac{7}{2}\right)\right)-5. \] Answer: \(y=\frac{4}{7}\sin\left(\frac{2\pi}{3}\left(x-\frac{7}{2}\right)\right)-5\)
10.
Question 10. Which of the following is equivalent to the expression \(\log_7(s)+8\log_7(w)+\log_7(z)\)?
Solution: Use the power rule: \[ 8\log_7(w)=\log_7(w^8) \] Then use the product rule: \[ \log_7(s)+\log_7(w^8)+\log_7(z)=\log_7(sw^8z) \] Answer: \(\log_7(sw^8z)\)
11.
Question 11. Evaluate \(\log_7(5764801)\).
Solution: We need to find the exponent that makes \(7\) equal to \(5764801\): \[ \log_7(5764801)=x \] means \[ 7^x=5764801. \] Since \[ 7^8=5764801, \] we have \[ \log_7(5764801)=8. \] Answer: 8
12.
Question 12. Solve \(\log_2 x=\log_2 6+\log_2 2\).
Solution: Use the product rule for logarithms: \[ \log_2 6+\log_2 2=\log_2(6\cdot 2) \] So \[ \log_2 x=\log_2 12. \] Therefore, \[ x=12. \] Answer: \(x=12\)
1 out of 1
