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Andreas93 [3]
3 years ago
11

Present Value Hoew much shuold be deposited in an account paying 7.8% interest compounded monthly in order to have a balance of

$21,000 after 4 years?
Mathematics
1 answer:
Alisiya [41]3 years ago
6 0

Answer: $15385 should be deposited.

Step-by-step explanation:

The principal was compounded monthly. This means that it was compounded 12 times in a year. So

n = 12

The rate at which the principal was compounded is 7.8%. So

r = 7.8/100 = 0.078

It was compounded for 4 years. Therefore,

t = 4

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years. The total amount is given as $21000. Therefore

21000 = P (1+0.078/12)^12×4

21000 = P (1+0.078/12)^48

21000 = P (1+0.0065)^48

21000 = P (1.0065)^48

P = 21000/1.365

P = $15385

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Use factoring to solve the quadratic equation x2 − 2x − 3 = 12.
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Can I get help solving this graph please?
Daniel [21]
see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100)  B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40)  C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0)  B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60)  C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then 
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x) 
</span>h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then 
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x) 
h'(x)=18/25=0.72 

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

<span> h'(x) = 1.44 ------------ > not exist</span>

8 0
3 years ago
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