Given:
The function are


To find:
The value of
.
Solution:
We have,


We need to find the value of
.
![[\because (f\circ g)(x)=f(g(x))]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28f%5Ccirc%20g%29%28x%29%3Df%28g%28x%29%29%5D)
![[\because g(x)=\sqrt{x}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20g%28x%29%3D%5Csqrt%7Bx%7D%5D)
![[\because f(x)=x^2-1]](https://tex.z-dn.net/?f=%5B%5Cbecause%20f%28x%29%3Dx%5E2-1%5D)
The value of
is 0.
Therefore, the correct option is A.
Answer:
It equals 3, but if you add two negatives inside of the parenthesis, then it equals -19.
Step-by-step explanation:
You have to plug in the numbers where the variables are, so it would be -8-(-11).
Answer:
Since every 30 days he wil have both lessons on the same day , and he already had both lessons on the last day of the previous month, that means that the day 30 the current month he wil have both lessons on the same day (It may be the last day if the month has 30 days or it may not be the last day if the month has 31 days)
Step-by-step explanation:
Lets find the least common factor of 5 and 6
Multiples of 5
5 10 15 20 35 30 35 40......
Multiples of 6
6 12 18 24 30 36
LCF of 5 and 6 = 30
Every 30 days he wil have both lessons on the same day
Answer:
0.85
Step-by-step explanation: