Ok, so remember that the derivitive of the position function is the velocty function and the derivitive of the velocity function is the accceleration function
x(t) is the positon function
so just take the derivitive of 3t/π +cos(t) twice
first derivitive is 3/π-sin(t)
2nd derivitive is -cos(t)
a(t)=-cos(t)
on the interval [π/2,5π/2) where does -cos(t)=1? or where does cos(t)=-1?
at t=π
so now plug that in for t in the position function to find the position at time t=π
x(π)=3(π)/π+cos(π)
x(π)=3-1
x(π)=2
so the position is 2
ok, that graph is the first derivitive of f(x)
the function f(x) is increaseing when the slope is positive
it is concave up when the 2nd derivitive of f(x) is positive
we are given f'(x), the derivitive of f(x)
we want to find where it is increasing AND where it is concave down
it is increasing when the derivitive is positive, so just find where the graph is positive (that's about from -2 to 4)
it is concave down when the second derivitive (aka derivitive of the first derivitive aka slope of the first derivitive) is negative
where is the slope negative?
from about x=0 to x=2
and that's in our range of being increasing
so the interval is (0,2)
I think it means you have to find the area of both shapes and then compare them to see if Liam is correct if he is not correct say why. Hope this helps :)
Answer:
x=13 and y=sqrt338
Step-by-step explanation:
so basically the opposite of one of the 45 degrees is 13 so the other 45 degrees is 13 too and you find pythagorean theorem of the other one
Answer:
Please check the explanation
Step-by-step explanation:
Given the function
Given that the output = -3
i.e. y = -3
now substituting the value y=-3 and solve for x to determine the input 'x'
switch sides
Add 1 to both sides
![\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}](https://tex.z-dn.net/?f=%5Cmathrm%7BFor%5C%3A%7Dg%5E3%5Cleft%28x%5Cright%29%3Df%5Cleft%28a%5Cright%29%5Cmathrm%7B%5C%3Athe%5C%3Asolutions%5C%3Aare%5C%3A%7Dg%5Cleft%28x%5Cright%29%3D%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%2C%5C%3A%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%5Cfrac%7B-1-%5Csqrt%7B3%7Di%7D%7B2%7D%2C%5C%3A%5Csqrt%5B3%5D%7Bf%5Cleft%28a%5Cright%29%7D%5Cfrac%7B-1%2B%5Csqrt%7B3%7Di%7D%7B2%7D)
Thus, the input values are:
![x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}](https://tex.z-dn.net/?f=x%3D-%5Csqrt%5B3%5D%7B2%7D%2B5%2C%5C%3Ax%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%5Cleft%281%2B5%5Ccdot%20%5C%3A2%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cright%29%7D%7B2%7D-i%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%5Csqrt%7B3%7D%7D%7B2%7D%2C%5C%3Ax%3D%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%5Cleft%281%2B5%5Ccdot%20%5C%3A2%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D%5Cright%29%7D%7B2%7D%2Bi%5Cfrac%7B%5Csqrt%5B3%5D%7B2%7D%5Csqrt%7B3%7D%7D%7B2%7D)
And the real input is:
![x=-\sqrt[3]{2}+5](https://tex.z-dn.net/?f=x%3D-%5Csqrt%5B3%5D%7B2%7D%2B5)
Answer:
y = -2
Step-by-step explanation:
Step 1: Combine like terms.
Step 2: Divide both sides by 9.
Step 3: Check if solution is correct.
Therefore, y = -2.
