Answer:
0.66071428571429
Step-by-step explanation:
Answer:
=
1
1
2
Step-by-step explanation:
.
Assume x are Erica's classes & y are Bo's classes.
x+y=35
x=2y-13
Replacing the value of x into the first equation
2y-13+y=35
3y=35+13
y=16 classes
x=2*16-13=19 classes
The value of sin-1(1) is negative startfraction pi over 2 endfraction, startfraction -pi over 2 endfraction.
<h3>What is the value of sin (π/2)?</h3>
The value of sin (π/2) is equal to the number 1. The value of the sin-1(1) has to be find out.
Suppose the value of this function is <em>x</em>. Thus,
![x=\sin^{-1}(1)](https://tex.z-dn.net/?f=x%3D%5Csin%5E%7B-1%7D%281%29)
Solve it further,
......1
The value of sin (π/2) and -sin (-π/2) is equal to 1 such that,
![\sin\dfrac{\pi}{2}=-\sin\left(-\dfrac{\pi}{2}\right)=1](https://tex.z-dn.net/?f=%5Csin%5Cdfrac%7B%5Cpi%7D%7B2%7D%3D-%5Csin%5Cleft%28-%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%29%3D1)
Put this value in the equation 1,
![\sin(x)=\sin\left(\dfrac{\pi}{2}\right)=-\sin\left(-\dfrac{\pi}{2}\right)](https://tex.z-dn.net/?f=%5Csin%28x%29%3D%5Csin%5Cleft%28%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%29%3D-%5Csin%5Cleft%28-%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%29)
Thus, the range will be,
![\left(\dfrac{\pi}{2},-\dfrac{\pi}{2}\right)](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7B%5Cpi%7D%7B2%7D%2C-%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%29)
Thus, the value of sin-1(1) is negative startfraction pi over 2 endfraction, startfraction -pi over 2 endfraction.
Learn more about the sine values here;
brainly.com/question/10711389