Answer:
1/9
Step-by-step explanation:
<span><span>(<span><span>2x</span>−9</span>)</span><span>(<span><span>4x</span>+1</span>)</span></span><span> =<span><span>(<span><span>2x</span>+<span>−9</span></span>)</span><span>(<span><span>4x</span>+1</span>)</span></span></span><span> =<span><span><span><span><span>(<span>2x</span>)</span><span>(<span>4x</span>)</span></span>+<span><span>(<span>2x</span>)</span><span>(1)</span></span></span>+<span><span>(<span>−9</span>)</span><span>(<span>4x</span>)</span></span></span>+<span><span>(<span>−9</span>)</span><span>(1)</span></span></span></span><span> =<span><span><span><span>8<span>x^2</span></span>+<span>2x</span></span>−<span>36x</span></span>−9</span></span><span> Answer: <span><span><span>8<span>x^2</span></span>−<span>34x</span></span>−<span>9</span></span></span>
Answer/Step-by-step explanation:
Recall the acronym for trigonometric ratios: SOHCAHTOA
SOH means sine of the reference angle = opp/hypotenuse
CAH means cosine of the reference angle = Adj/hyp
TOA means tangent of the reference angle = opp/adj
a. Reference angle = A
opp = 15
hypotenuse = 17
Therefore:
![sin(A) = \frac{opp}{hypo} = \frac{15}{17}](https://tex.z-dn.net/?f=%20sin%28A%29%20%3D%20%5Cfrac%7Bopp%7D%7Bhypo%7D%20%3D%20%5Cfrac%7B15%7D%7B17%7D%20)
b. Reference angle = A
![cos(A) = \frac{adj}{hypo}](https://tex.z-dn.net/?f=%20cos%28A%29%20%3D%20%5Cfrac%7Badj%7D%7Bhypo%7D%20)
Adj = 8
hypotenuse = 17
Therefore:
![cos(A) = \frac{8}{17}](https://tex.z-dn.net/?f=%20cos%28A%29%20%3D%20%5Cfrac%7B8%7D%7B17%7D%20)
c. Reference angle = A
![tan(A) = \frac{opp}{adj}](https://tex.z-dn.net/?f=%20tan%28A%29%20%3D%20%5Cfrac%7Bopp%7D%7Badj%7D%20)
Adj = 8
Opp = 15
Therefore:
![tan(A) = \frac{15}{8}](https://tex.z-dn.net/?f=%20tan%28A%29%20%3D%20%5Cfrac%7B15%7D%7B8%7D%20)
d. Reference angle = B
![sin(B) = \frac{opp}{hypo}](https://tex.z-dn.net/?f=%20sin%28B%29%20%3D%20%5Cfrac%7Bopp%7D%7Bhypo%7D%20)
Opp = 8
hypotenuse = 17
Therefore:
![sin(B) = \frac{8}{17}](https://tex.z-dn.net/?f=%20sin%28B%29%20%3D%20%5Cfrac%7B8%7D%7B17%7D%20)
e. Reference angle = B
![cos(B) = \frac{adj}{hypo}](https://tex.z-dn.net/?f=%20cos%28B%29%20%3D%20%5Cfrac%7Badj%7D%7Bhypo%7D%20)
Adj = 15
hypotenuse = 17
Therefore:
![cos(B) = \frac{15}{17}](https://tex.z-dn.net/?f=%20cos%28B%29%20%3D%20%5Cfrac%7B15%7D%7B17%7D%20)
d. Reference angle = B
![tan(B) = \frac{opp}{adj}](https://tex.z-dn.net/?f=%20tan%28B%29%20%3D%20%5Cfrac%7Bopp%7D%7Badj%7D%20)
Opp = 8
Adj = 15
Therefore:
![tan(B) = \frac{8}{15}](https://tex.z-dn.net/?f=%20tan%28B%29%20%3D%20%5Cfrac%7B8%7D%7B15%7D%20)
Answer:
2x² + 3x + 8 + 7/x-4
Step-by-step explanation:
For this problem, we can use synthetic division.
4 | 2 -5 -4 -25
| 8 12 32
|_________
2 3 8 7/x-4
This means the quotient is 2x² + 3x + 8 + 7/x-4.