Answer:
The answer is A. (4x²+3)(x+2)
Step-by-step explanation:
If u multiply (4x²+3) by (x+2), u get 4x^3+8x^2+3x+6
Multiply 5% by 229
So, .05 x 229 = 11.45
$11.45 tax
Answer:
7. 10 8. 20 9. 100 thats all i know
Step-by-step explanation:
Let

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

So, the base case is ok. Now, we need to assume
and prove
.
states that

Since we're assuming
, we can substitute the sum of the first n terms with their expression:

Which terminates the proof, since we showed that

as required