![\bf y=x+2cos(x)\qquad \qquad [-2\pi ,2\pi ]\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=1-2sin(x)\implies 0=1-2sin(x)\implies sin(x)=\cfrac{1}{2} \\\\\\ \measuredangle x=sin^{-1}\left( \frac{1}{2} \right)\implies \measuredangle x= \begin{cases} \frac{\pi }{6}\\\\ \frac{5\pi }{6} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20y%3Dx%2B2cos%28x%29%5Cqquad%20%5Cqquad%20%5B-2%5Cpi%20%2C2%5Cpi%20%5D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1-2sin%28x%29%5Cimplies%200%3D1-2sin%28x%29%5Cimplies%20sin%28x%29%3D%5Ccfrac%7B1%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cmeasuredangle%20x%3Dsin%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B1%7D%7B2%7D%20%5Cright%29%5Cimplies%20%5Cmeasuredangle%20x%3D%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7B%5Cpi%20%7D%7B6%7D%5C%5C%5C%5C%0A%5Cfrac%7B5%5Cpi%20%7D%7B6%7D%0A%5Cend%7Bcases%7D)
only in the 1st and 2nd quadrants, where sine is positive
We have the following equation:
<span> h(t)=-4.92t^2+17.69t+575
</span> For the domain we have:
<span> </span>We match zero:
-4.92t ^ 2 + 17.69t + 575 = 0
We look for the roots:
t1 = -9.16
t2 = 12.76
We are left with the positive root, so the domain is:
[0, 12.76]
For the range we have:
We derive the function:
h '(t) = - 9.84t + 17.69
We equal zero and clear t:
-9.84t + 17.69 = 0
t = 17.69 / 9.84
t = 1.80
We evaluate the time in which it reaches the maximum height in the function:
h (1.80) = - 4.92 * (1.80) ^ 2 + 17.69 * (1.80) +575
h (1.80) = 590.90
Therefore, the range is given by:
[0, 590.9]
Answer:
the domain and range are:
domain: [0, 12.76] range: [0, 590.9]
Answer:
the answer is 5
Step-by-step explanation:
hope that helps
Answer:
331
Step-by-step explanation:
You just divide 3,972 by 12.
Answer:
x=15in-4/3in
i=8/-6nx-30n
n=-8/6ix+30i
Step-by-step explanation: