I'm assuming all of (x^2+9) is in the denominator. If that assumption is correct, then,
One possible answer is 
Another possible answer is 
There are many ways to do this. The idea is that when we have f( g(x) ), we basically replace every x in f(x) with g(x)
So in the first example above, we would have
In that third step, g(x) was replaced with x^2+9 since g(x) = x^2+9.
Similar steps will happen with the second example as well (when g(x) = x^2)
Answer:
Step-by-step explanation:
We translate the phrase into an equation:
Then we simplify:
Answer:
1, yes add one, 2, no, 3, yes, you add 10. 4, no, 5, yes, you add 2.
Step-by-step explanation:
1 add one, 3 add 10 5 add 2.
Your welcome :)
Answer:the correct answer should be d.2.07
Step-by-step explanation:
Answer:
Step-by-step explanation:
Polynomial f(x) has the following conditions: zeros of -2 (multiplicity 3), 3 (multiplicity 1), and with f(0) = 120.
The first part zeros of -2 means (x+2) and multiplicity 3 means (x+2)^3.
The second part zeros of 3 means (x-3) and multiplicity 1 means (x-3).
The third part f(0) = 120 means substituting x=0 into (x+2)^3*(x-3)*k =120
(0+2)^3*(0-3)*k = 120
-24k = 120
k = -5
Combining all three conditions, f(x)
= -5(x+2)^3*(x-3)
= -5(x^3 + 3*2*x^2 + 3*2*2*x + 2^3)(x-3)
= -5(x^4 + 6x^3 + 12x^2 + 8x - 3x^3 - 18x^2 - 36x - 24)
= -5(x^4 + 3x^3 - 6x^2 - 28x -24)
= -5x^4 - 15x^3 + 30x^2 + 140x + 120
