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LUCKY_DIMON [66]
3 years ago
5

The sum of two #'s is 29. The sum of the smaller and 2 times the larger is 45. Find the #'s.

Mathematics
2 answers:
podryga [215]3 years ago
6 0

x+y=29

y=29-x

x+2y=45

x+2(29-x)=45

x+58-2x=45

-1x=-13

x=13

y=29-13=16

x+y = 13+16 = 29

x+2y= 13 + 2(16) = 13+32 = 45

so the numbers are 13 & 16

Solnce55 [7]3 years ago
3 0
X= 13
y=16

Work:

x+y=29

x=29-y

x+2y=45

29-y+2y=45

y=45-29

y=16

x=29-16

x=13
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4 years ago
David’s bank offers a 36-month Certificate of Deposit (CD) with an APR of 2.25%. Use the compound interest formula to answer the
tamaranim1 [39]

Using compound interest, it is found that:

a) A(8) = 2389.66

b) t = 31.15

c) P = 1870.85

Compound interest:

A(t) = P\left(1 + \frac{r}{n}\right)^{nt}

  • A(t) is the amount of money after t years.  
  • P is the principal(the initial sum of money).  
  • r is the interest rate(as a decimal value).  
  • n is the number of times that interest is compounded per year.  
  • t is the time in years for which the money is invested or borrowed.

In this problem:

  • The APR is of 2.25%, hence r = 0.0225.
  • No information about the number of compounding per year, hence n = 1.

Item a:

P = 2000, hence:

A(t) = P\left(1 + \frac{r}{n}\right)^{nt}

A(8) = 2000\left(1 + \frac{0.0225}{1}\right)^{8}

A(8) = 2389.66

Item b:

A(t) = 4000, hence:

A(t) = P\left(1 + \frac{r}{n}\right)^{nt}

4000 = 2000\left(1 + \frac{0.0225}{1}\right)^{t}

(1.0225)^t = 2

\log{(1.0225)^t} = \log{2}

t\log{1.0225} = \log{2}

t = \frac{\log{2}}{\log{1.0225}}

t = 31.15

Item c:

A(3) = 2000, hence:

A(t) = P\left(1 + \frac{r}{n}\right)^{nt}

2000 = P\left(1 + \frac{0.0225}{1}\right)^{3}

P = \frac{2000}{(1.0225)^3}

P = 1870.85

A similar problem is given at brainly.com/question/24850750

7 0
2 years ago
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