Answer:
B and C only show direct proportion
The correct answer should be B.
keeping in mind that anything raised at the 0 power, is 1, with the sole exception of 0 itself.
![\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{(r^{-7}b^{-8})^0}{t^{-4}w}\implies \cfrac{1}{t^{-4}w}\implies \cfrac{1}{t^{-4}}\cdot \cfrac{1}{w}\implies t^4\cdot \cfrac{1}{w}\implies \cfrac{t^4}{w}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bnegative%20exponents%7D%0A%5C%5C%5C%5C%0Aa%5E%7B-n%7D%20%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5En%7D%0A%5Cqquad%20%5Cqquad%0A%5Ccfrac%7B1%7D%7Ba%5En%7D%5Cimplies%20a%5E%7B-n%7D%0A%5Cqquad%20%5Cqquad%20a%5En%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5E%7B-n%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%28r%5E%7B-7%7Db%5E%7B-8%7D%29%5E0%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7D%7D%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20t%5E4%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20%5Ccfrac%7Bt%5E4%7D%7Bw%7D%20)
Answer:
-1
Step-by-step explanation:
Any variable without a visible coefficient means it has a coefficient of 1. Since it's negated, the coefficient is -1.
Recall the sum identity for cosine:
cos(a + b) = cos(a) cos(b) - sin(a) sin(b)
so that
cos(a + b) = 12/13 cos(a) - 8/17 sin(b)
Since both a and b terminate in the first quadrant, we know that both cos(a) and sin(b) are positive. Then using the Pythagorean identity,
cos²(a) + sin²(a) = 1 ⇒ cos(a) = √(1 - sin²(a)) = 15/17
cos²(b) + sin²(b) = 1 ⇒ sin(b) = √(1 - cos²(b)) = 5/13
Then
cos(a + b) = 12/13 • 15/17 - 8/17 • 5/13 = 140/221
