![\bf cos\left[tan^{-1}\left(\frac{12}{5} \right)+ tan^{-1}\left(\frac{-8}{15} \right) \right]\\ \left. \qquad \qquad \quad \right.\uparrow \qquad \qquad \qquad \uparrow \\ \left. \qquad \qquad \quad \right.\alpha \qquad \qquad \qquad \beta \\\\\\ \textit{that simply means }tan(\alpha)=\cfrac{12}{5}\qquad and\qquad tan(\beta)=\cfrac{-8}{5} \\\\\\ \textit{so, we're really looking for }cos(\alpha+\beta)](https://tex.z-dn.net/?f=%5Cbf%20cos%5Cleft%5Btan%5E%7B-1%7D%5Cleft%28%5Cfrac%7B12%7D%7B5%7D%20%20%5Cright%29%2B%20tan%5E%7B-1%7D%5Cleft%28%5Cfrac%7B-8%7D%7B15%7D%20%20%5Cright%29%20%5Cright%5D%5C%5C%0A%5Cleft.%20%5Cqquad%20%20%5Cqquad%20%20%5Cquad%20%20%20%5Cright.%5Cuparrow%20%5Cqquad%20%5Cqquad%20%20%5Cqquad%20%20%5Cuparrow%20%5C%5C%0A%5Cleft.%20%5Cqquad%20%20%5Cqquad%20%20%5Cquad%20%20%20%5Cright.%5Calpha%20%5Cqquad%20%5Cqquad%20%20%5Cqquad%20%20%5Cbeta%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bthat%20simply%20means%20%7Dtan%28%5Calpha%29%3D%5Ccfrac%7B12%7D%7B5%7D%5Cqquad%20and%5Cqquad%20tan%28%5Cbeta%29%3D%5Ccfrac%7B-8%7D%7B5%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20we%27re%20really%20looking%20for%20%7Dcos%28%5Calpha%2B%5Cbeta%29)
now.. hmmm -8/15 is rather ambiguous, since the negative sign is in front of the rational, and either 8 or 15 can be negative, now, we happen to choose the 8 to get the minus, but it could have been 8/-15
ok, well hmm so, the issue boils down to
now, let's take a peek at the second angle, angle β
now, with that in mind, let's use the angle sum identity for cosine
If you’re looking for the volume of a sphere, it is
V = 4/3πr³
Answer:
1. I assume that stretching vertically by factor of 2 means that every point on the transformed curve is twice as far from the x axis as the original curve.
Imagine that you have plotted y=5x. you want to change the y scale so that each point on the original curve is two times as high. For example, one point on the original curve is (1,5). This point must be relocated to (1,10). Similarly, the point (2,10) on the original curve must be relocated to (2,20).
You can see that these relocations are accomplished if we transform the function to y=2(5x)=10x.
Step-by-step explanation:
hope this helps! even though it's just the first one...
Answer: 2.58
Step-by-step explanation:
Answer:
Law of Cosines
Angle A = 45°
Angle B = 51°
Angle C = 84°
Step-by-step explanation:
Law of sines is used when we are given
a) two angles and one side or
b) two sides and non-included angle
Law of cosines is used when we are given
a) three sides or
b) two sides and included angle
In the given question we are given three sides so, Law of Cosines will be used to solve the triangle.
Law of Cosines is:
We will find the three angles A ,B and C of the triangle using above formula.
a= 10, b=11, c=14
Putting values and finding angle A
Now finding angle B
Now finding angle C
