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:
(20 divided by 4) is the quotient
Step-by-step explanation:
Answer: 
<u>Step-by-step explanation:</u>
Pythagorean Theorem is: a² + b² = c² , <em>where "c" is the hypotenuse</em>
<em />

Note: 4² + (8√2)² = hypotenuse² → hypotenuse = 12

Note: 12² + opposite² = 20² → opposite = 16

Note: adjacent² + 5² = 6² → adjacent = √11

Note: adjacent² + 7² = (13√2)² → adjacent = 17
see the attached figure to better understand the problem
we know that
The volume of the cone is equal to

in this problem

Substitute the values in the formula above

therefore
the answer is
The volume of the nose cone is 
Answer:
MArginal productivity: 
We can interpret this as he will reduce his time an <em>additional </em>0.0002 seconds for every <em>additional </em>yard he trains.
Step-by-step explanation:
The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.
In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.
The marginal productivity can be expressed as:

where dt is the variation in time and dL is the variation in training yards.
We can not derive the function because it is not defined, but we can approximate with the last two points given:

Then we can interpret this as he will reduce his time an <em>additional </em>0.0002 seconds for every <em>additional </em>yard he trains.
This is an approximation that is valid in the interval of 60,000 to 70,000 yards of training.