Depends if you want
1. find how much he will earn, find the differnce between that and 18000
2. see how much to invest till he will get 18000
A=

A=futre amount
P=present amout
r=rate in decimal
n=number of times per year ccompounded
t=time in years
1.
A=?
P=9000
r=0.06
n=4 (quarter means 4 times per year)
t=2
?=

?=

?=

?=10138.4 will be earned
18000-10138.4=7861.6 needed
2.
A=18000
P=9000+x
r=0.06
n=4 (quarter means 4 times per year)
t=2
18000=

18000=

18000=

divide both sides by 1.015^8
15978.8=9000+x
minus 9000 both sides
6978.8 needed
if he willnot be investing any more, he needs $7861.6 more
if he will invest more he will need to invest $6978.8 more
Answer:
the answer is 52
Step-by-step explanation:
you have to divide 468 by 9 and should get you too 52
A = (1/2) * b * h
The base is 10; the x-value going from 0 to 10.
The perpendicular height is 2; the y-value going from 0 to 2
A = (1/2) * 10 * 2
A = 10 units ^2
Answer:
If s(x) = x – 7 and t(x) = 4x2 – x + 3, which expression is equivalent to (t*s)(x)? 4(x – 7)2 – x – 7 + 3 4(x – 7)2 – (x – 7) + 3 (4x2 – x + 3) – 7 (4x2 – x + 3)(x – 7)
Answer: C) For every original price, there is exactly one sale price.
For any function, we always have any input go to exactly one output. The original price is the input while the output is the sale price. If we had an original price of say $100, and two sale prices of $90 and $80, then the question would be "which is the true sale price?" and it would be ambiguous. This is one example of how useful it is to have one output for any input. The input in question must be in the domain.
As the table shows, we do not have any repeated original prices leading to different sale prices.