9514 1404 393
Answer:
A ≈ 65.4°
Step-by-step explanation:
The mnemonic SOH CAH TOA is intended to remind you of the trig relations associated with pairs of sides. For angle A, given sides b and c are the adjacent side and the hypotenuse, respectively. The appropriate relation is ...
Cos = Adjacent/Hypotenuse
cos(A) = b/c
A = arccos(b/c) = arccos(2.5/6)
A ≈ 65.4°
By the binomial theorem, the
term in the expansion of
is
![\dbinom n21^{n-2}\left(-\dfracx4\right)^2=\dfrac{n!}{2!(n-2)!}\dfrac{x^2}{16}](https://tex.z-dn.net/?f=%5Cdbinom%20n21%5E%7Bn-2%7D%5Cleft%28-%5Cdfracx4%5Cright%29%5E2%3D%5Cdfrac%7Bn%21%7D%7B2%21%28n-2%29%21%7D%5Cdfrac%7Bx%5E2%7D%7B16%7D)
which suggests that the contribution of the binomial coefficient should make up the remaining factor of 21. That is,
![\dfrac{n!}{2!(n-2)!}=\dfrac{n(n-1)}2=21\implies n=7](https://tex.z-dn.net/?f=%5Cdfrac%7Bn%21%7D%7B2%21%28n-2%29%21%7D%3D%5Cdfrac%7Bn%28n-1%29%7D2%3D21%5Cimplies%20n%3D7)
<span>Let's try to solve the equation:
1/x + 1/(x)² = 2
Kelly says that it is not possible because there are the variable x and x² in the denominators. Kelly is correct in that there is a value of x that makes the denominator zero. In this case, x = 0 makes the denominator of 1/x zero and also makes the denominator of 1/x² = 0.
</span>But, we want to look for values of x that will make the whole equation true, not the values of x that make the denominators zero. 1/x + 1/(x)² = 2
(x +1)/(x)² = 2
Multiply through by x² with the proviso that x is not 0.
Then,
(x + 1) = 2x²
At this point, we are looking for solutions to (x + 1) = 2x² which is related to but not identical to the original equation. So, we will have to check any answers we get to
(x + 1) = 2x² against the original problem: 1/x + 1/(x)² = 2
I think it’s B but I’m not sure
Answer:
![\sin \frac{3\pi}{2} = -1](https://tex.z-dn.net/?f=%5Csin%20%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D%20-1)
![\cos\frac{3\pi}{2} =0](https://tex.z-dn.net/?f=%5Ccos%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D0)
Step-by-step explanation:
Please refer to the image attached.
Here we have a circle with unit radius. At some angle Ф the radius = 1 , and it is the hypotenuse (shown by green line in the image attached) of the ΔPQR thus formed. As our angle Ф increases, the hypotenuse gets closer to the positive y axis and at 90°, it overlap the y axis. Hypotenuse (H) and Opposite site (O) becomes same and Adjacent (A) becomes 0.
As our angle move further and reaches 180, the Hypotenuse and adjacent becomes same and overlap negative x axis. As we move further at 270 i.e
, the hypotenuse and opposite side overlap on y axis and Adjacent side become 0. However the opposite side becomes negative here .
Our sine ratio says
![\sin \frac{3\pi}{2} =\frac{opposite}{Hypotenuse}](https://tex.z-dn.net/?f=%5Csin%20%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D%5Cfrac%7Bopposite%7D%7BHypotenuse%7D)
![\sin \frac{3\pi}{2} =\frac{-1}{1}](https://tex.z-dn.net/?f=%5Csin%20%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D%5Cfrac%7B-1%7D%7B1%7D)
![\sin \frac{3\pi}{2} =-1](https://tex.z-dn.net/?f=%5Csin%20%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D-1)
Hence we have our ![\sin \frac{3\pi}{2} = -1](https://tex.z-dn.net/?f=%5Csin%20%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D%20-1)
Now
![\cos\frac{3\pi}{2} =\frac{Adjacent}{Hypotenuse}](https://tex.z-dn.net/?f=%5Ccos%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D%5Cfrac%7BAdjacent%7D%7BHypotenuse%7D)
Adjacent as we discussed is 0 at ![\frac{3\pi}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B3%5Cpi%7D%7B2%7D)
![\cos\frac{3\pi}{2} =\frac{0}{1}](https://tex.z-dn.net/?f=%5Ccos%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D%5Cfrac%7B0%7D%7B1%7D)
![\cos\frac{3\pi}{2} =0](https://tex.z-dn.net/?f=%5Ccos%5Cfrac%7B3%5Cpi%7D%7B2%7D%20%3D0)