EBUS CAL12 Chapter 1 Quiz 2-4 (Lesson 5,6,7)

Solution: For a cosine function in the form \(y=a\cos(b(x-c))+d\), the amplitude is \(|a|\). Here, \(a=4\), so the amplitude is \(|4|=4\). Answer: A. True

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: D. 3

Solution: The slope-intercept form is \(y=mx+b\), where \(m\) is the slope. For \(y=2x+b\), the coefficient of \(x\) is \(2\). Therefore, every line in this family has slope \(2\). Answer: B. Their slope is 2.

Solution: Let the inverse function be \(y=\frac{x-6}{3}\). To find the original function, switch \(x\) and \(y\): \[x=\frac{y-6}{3}\] Solve for \(y\): \[3x=y-6\] \[y=3x+6\] Answer: D. \(3x+6\)

Solution: \(\cos^{-1}\left(\frac{1}{2}\right)\) asks for the angle whose cosine is \(\frac{1}{2}\). From special angles, \(\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}\). Therefore, \(\cos^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\). Answer: B. True

Solution: Rewrite \(27\) as a power of \(3\): \[27=3^3\] Therefore, \[\log_3 27=\log_3(3^3)=3\] Answer: B. True

Solution: Use the power rule for logarithms: \[-\frac{1}{2}\log(x)=\log\left(x^{-\frac{1}{2}}\right)=\log\left(\frac{1}{\sqrt{x}}\right)\] Also, \[2\log(3)=\log(3^2)=\log(9)\] Now combine using the product rule: \[\log\left(\frac{1}{\sqrt{x}}\right)+\log(9)=\log\left(\frac{9}{\sqrt{x}}\right)\] Answer: C. \(\log\left(\frac{9}{\sqrt{x}}\right)\)

Solution: Find the slope: \[m=\frac{4-(-1)}{-4-(-2)}=\frac{5}{-2}=-\frac{5}{2}\] Use slope-intercept form \(y=mx+b\). Substitute \((-2,-1)\): \[-1=-\frac{5}{2}(-2)+b\] \[-1=5+b\] \[b=-6\] Therefore, the equation is \[y=-\frac{5}{2}x-6\] Answer: A. \(y=-\frac{5}{2}x-6\)

Solution: Use the transformed sine form: \[y=a\sin(b(x-c))+d\] The amplitude is \(|a|=\frac{5}{3}\), so \(a=\frac{5}{3}\). The period is \(\frac{2\pi}{b}\). Since the period is \(\frac{1}{2}\), \[\frac{2\pi}{b}=\frac{1}{2}\] \[b=4\pi\] A phase shift \(\frac{1}{2}\) rad to the right means \(x-\frac{1}{2}\). A vertical translation \(5\) units down means \(-5\). Thus, \[y=\frac{5}{3}\sin\left(4\pi\left(x-\frac{1}{2}\right)\right)-5\] Answer: D. \(y=\frac{5}{3}\sin\left(4\pi\left(x-\frac{1}{2}\right)\right)-5\)

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: D. \(\log_7(sw^8z)\)

Solution: Rewrite \(32768\) as a power of \(8\): \[8^5=32768\] Therefore, \[\log_8 32768=5\] Answer: D. 5

Solution: Use the product rule: \[\log_3 2+\log_3 3=\log_3(2\cdot 3)=\log_3 6\] So, \[\log_3 x=\log_3 6\] Since the logarithms have the same base, \[x=6\] Answer: B. \(x=6\)

Solution: Start with \[57=2^x\] Take logarithms of both sides: \[\log(57)=\log(2^x)\] Use the power rule: \[\log(57)=x\log(2)\] Solve for \(x\): \[x=\frac{\log(57)}{\log(2)}\] Calculate: \[x\approx 5.8\] Answer: D. 5.8