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krek1111 [17]
3 years ago
6

Can someone solve 5 + 2x = + 6 and n - 3n = 14 - 4n and 7 ( 5a - 4 ) - 1 = 14 - 8a

Mathematics
2 answers:
Vladimir79 [104]3 years ago
8 0

Step-by-step explanation:

1.

5 + 2x = + 6

2x=6-5

2x=1

x \ =  \frac{1}{2}

2.

n - 3n = 14 - 4n

-2n= 14 -4n

-2n+4n=14

2n=14

n=7

3.

7(5a - 4 ) - 1 = 14 - 8a

35a-28-1=14-8a

35a-29=14-8a

35a+8a=14+29

43a=43

a= 1

professor190 [17]3 years ago
6 0

Answer:

5 + 2x = 6

2x = 6-5

2x = 1

x = 0.5

n -3n = 14 -4n

-2n = 14-4n (since 4n is negative, change the sign to make it positive and then add 4n to both sides of the equation)

2n = 14

n = 7

7 (5a - 4) - 1 = 14 - 8a

35a - 28 - 1 = 14 -8a

35a -29 = 14 - 8a

43a = 43

a = 1

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Part b) \$20,442.36

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Step-by-step explanation:

we know that  

The compound interest formula is equal to

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

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest  in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part a) How much would you have at the end of 1 year?

in this problem we have

t=1\ years\\ P=\$15,192\\ r=0.16\\n=1

substitute in the formula above

A=15,192(1+\frac{0.16}{1})^{1*1}=\$17,622.72

Part b) How much would you have at the end of 2 year?

in this problem we have

t=2\ years\\ P=\$15,192\\ r=0.16\\n=1

substitute in the formula above

A=15,192(1+\frac{0.16}{1})^{1*2}=\$20,442.36

Part c) How much would you have at the end of 3 year?

in this problem we have

t=3\ years\\ P=\$15,192\\ r=0.16\\n=1

substitute in the formula above

A=15,192(1+\frac{0.16}{1})^{1*3}=\$23,713.13

Part d) How much would you have at the end of 4 year?

in this problem we have

t=4\ years\\ P=\$15,192\\ r=0.16\\n=1

substitute in the formula above

A=15,192(1+\frac{0.16}{1})^{1*4}=\$27,507.23

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3 years ago
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