Answer:
not possible, bother terms have to contain a letter variable
The answer is 13 you have to add the top and bottom
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)
Answer:
The given equation cannot be written in slope intercept form
Step-by-step explanation:
we know that
Is the equation of the line in slope intercept form
where
m is the slope
b is the y-intercept
In this problem we have
This is a vertical line (parallel to the y-axis)
The slope is undefined
therefore
The given equation cannot be written in slope intercept form
