I think is Y=5x+20
Hope you get it right!
:D
Answer:
We have function,

Standard Form of Sinusoid is

Which corresponds to

where a is the amplitude
2pi/b is the period
c is phase shift
d is vertical shift or midline.
In the equation equation, we must factor out 2 so we get

Also remeber a and b is always positive
So now let answer the questions.
a. The period is


So the period is pi radians.
b. Amplitude is

Amplitude is 6.
c. Domain of a sinusoid is all reals. Here that stays the same. Range of a sinusoid is [-a+c, a-c]. Put the least number first, and the greatest next.
So using that<em> rule, our range is [6+3, -6+3]= [9,-3] So our range</em> is [-3,9].
D. Plug in 0 for x.





So the y intercept is (0,-3)
E. To find phase shift, set x-c=0 to solve for phase shift.


Negative means to the left, so the phase shift is pi/4 units to the left.
f. Period is PI, so use interval [0,2pi].
Look at the graph above,
Answer:
I think the function is that she is working.
Answer:
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C
Step-by-step explanation:
∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗
Let 1st=arctan(x)
And 2nd=1
∫▒〖arctan(x).1 dx=arctan(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗
As we know that
derivative of arctan(x)=1/(1+x^2 )
∫▒〖1 dx〗=x
So
∫▒〖arctan(x).1 dx=arctan(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1
Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now
Let 1+x^2=u
du=2xdx
Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get
1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)
1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)
1/2 ∫▒(2xdx/u) =1/2 ln(u)+C
1/2 ∫▒(2xdx/u) =1/2 ln(1+x^2 )+C
Putting values in Eq1 we get
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C (required soultion)