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

Solution: For a cosine function in the form \(y=a\cos(bx+c)\), the period is \(\frac{2\pi}{|b|}\). Here, \(b=2\), so \[\text{Period}=\frac{2\pi}{2}=\pi.\] Therefore, the statement is true. Answer: A. True

Solution: The equation \(y=2x+b\) is in slope-intercept form \(y=mx+b\). The slope is \(m=2\), and \(b\) can change. So all members of this family of lines have slope 2. Answer: C. Their slope is 2.

Solution: The degree of a polynomial is the highest power of \(x\). In \(f(x)=5x-4x^2-x^3\), the highest power is \(3\). Therefore, the degree is 3. Answer: B. 3

Solution: Let \(y=x^7\). Switch \(x\) and \(y\): \[x=y^7.\] Solve for \(y\): \[y=\sqrt[7]{x}=x^{\frac{1}{7}}.\] Therefore, the inverse is \(f^{-1}(x)=x^{\frac{1}{7}}\). Answer: B. \(x^{\frac{1}{7}}\)

Solution: Find the inverse of \(y=\frac{1}{x^3}\). Switch \(x\) and \(y\): \[x=\frac{1}{y^3}.\] Then \[xy^3=1,\] \[y^3=\frac{1}{x},\] \[y=\sqrt[3]{\frac{1}{x}}=x^{-\frac{1}{3}}.\] This matches \(g(x)=x^{-\frac{1}{3}}\), so the statement is true. Answer: A. True

Solution: Rewrite \(16\) as a power of 2: \[16=2^4.\] So \[2^{-x}=2^4.\] Since the bases are equal, \[-x=4.\] Therefore, \[x=-4.\] The statement is true. Answer: A. True

Solution: Use the change of base formula: \[\log_4(120)=\frac{\log(120)}{\log(4)}.\] Calculate: \[\log_4(120)\approx 3.45.\] So the statement is true. Answer: A. True

Solution: First find the slope: \[m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-9}{6-3}=-\frac{5}{3}.\] Use \(y=mx+b\) and substitute \((3,9)\): \[9=-\frac{5}{3}(3)+b.\] \[9=-5+b.\] \[b=14.\] Therefore, the equation is \[y=-\frac{5}{3}x+14.\] Answer: A. \(y=-\frac{5}{3}x+14\)

Solution: Use the form \[y=a\sin(b(x-c))+d.\] The amplitude is \(\frac{4}{7}\), so \(a=\frac{4}{7}\). The period is 3, so \[b=\frac{2\pi}{\text{period}}=\frac{2\pi}{3}.\] 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: D. \(y=\frac{4}{7}\sin\left(\frac{2\pi}{3}\left(x-\frac{7}{2}\right)\right)-5\)

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: B. \(\log_9(rw^4x)\)

Solution: We need to find the exponent that makes \(6\) equal to \(279936\): \[6^7=279936.\] Therefore, \[\log_6(279936)=7.\] Answer: A. 7

Solution: Use the product rule of logarithms: \[\log_2(6)+\log_2(2)=\log_2(12).\] So \[\log_2(x)=\log_2(12).\] Therefore, \[x=12.\] Answer: C. \(x=12\)

Solution: Take logarithms of both sides: \[13=7^x.\] \[x=\log_7(13).\] Use the change of base formula: \[x=\frac{\log(13)}{\log(7)}.\] Calculate: \[x\approx 1.3.\] Answer: 1.3