let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B6%7D%7D%5Cqquad%20%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B6%5E2-5%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B36-25%7D%5Cimplies%20%5Cpm%20%5Csqrt%7B11%7D%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Type it into ur calculator as √175 and if you need radical form it’ll be 5√7 or just 13.2 (13.2287565553)
Answer:
Step-by-step explanation:
or
Step-by-step explanation:
has:
1) amplitude=|a|
2) phase shift=b
3) period=2pi/c
d) vertical shift=d
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You are given the amplitude is 2. This means a could either be 2 or -2.
You are given phase shift is -1/8. This means b is -1/8.
You are given the period is pi. So we need to solve pi=2pi/c.
pi=2pi/c
Dividing both sides by pi gives:
1=2/c
Multiply both sides by c gives:
c=2
The vertical shift is -2 so d=-2.
The equation could either be:
y=2cos(2(x-(-1/8)))-2
or
y=-2cos(2(x-(-1/8)))-2
Simplifying a little gives:
y=2cos(2x+1/4)-2
y=-2cos(2x+1/4)-2
Well i don't know if this will help but slope intercept form
y=mx+b
M-slope
B-Y intercept
and
(x,y)- would be any point on the graph and if you were to plug those points in for x and y the equation would solve if done properly
Answer:
Omg its totally 726263893276273
