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
360 - 250 = 110
m<1 = 110/2 = 55
answer
55
Answer:
B. cos−1(StartFraction 11.9 Over 14.5 EndFraction) = θ
Step-by-step explanation:
From definition:
cos(θ) = adjacent/hypotenuse
The adjacent side respect angle GFE (or θ) is side FE, and side FG is the hypotenuse. Replacing with data and isolating θ:
cos(θ) = 11.9/14.5
θ = cos^-1(11.9/14.5)
Gallons per day.......gallons / day.....so u put the gallons over the number of days, then divide
(1/2) / 6 = 1/2 * 1/6 = 1/12 of a gallon per day <==
If there is such a scalar function <em>f</em>, then



Integrate both sides of the first equation with respect to <em>x</em> :

Differentiate both sides with respect to <em>y</em> :


Integrate both sides with respect to <em>y</em> :

Plug this into the equation above with <em>f</em> , then differentiate both sides with respect to <em>z</em> :



Integrate both sides with respect to <em>z</em> :

So we end up with
