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kupik [55]
2 years ago
10

Heather purchased a new $1,500 laptop by using a credit card with a 17% interest rate. She will pay off the balance in 2 years b

y paying monthly payments of $74.16 . Calculate Heathers total cost of repayment.
Mathematics
1 answer:
olga2289 [7]2 years ago
7 0

Answer:

1779.84

Step-by-step explanation:

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If x divided 9=9 how do u know what x is
Volgvan
X is 1 because if you times it by x it equals it number
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3 years ago
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The weight of a box varies directly as the volume of the box. If a 138-pound box has a volume of 23 gallons, what is the weight
kolezko [41]

Answer:

D. 150 pounds

Step-by-step explanation:

For be directly proportional, it has to be that as one value increases the other does it equally, so

If  138 pound box has a volume of 23 gallons for 25 gallons ?. For being directly proportional tha value must be higest

138  →  23                       x=\frac{25*138}{23} = 150

x      →  25

5 0
3 years ago
How to solve 8=3/4w on math
NeTakaya
When you have to divide by a fraction, you have to multiply the other number by the fraction's reciprocal.

8 = 3/4 w    Divide by 3/4

w = 8/(3/4)

w = 8/ (3/4)  =  8 x 4/3   Multiply by the numerator and divide by the                                                   denominator.

w = (8 x 4) / 3

w = 32 / 3

w = 10 2/3
6 0
3 years ago
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Please help:((
kolezko [41]

Answer:

1/8 = 0.125

3/5= 0.60

30%=0.3

0.45=9/20

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Hey try to do these question by your self next time OwO

4 0
3 years ago
Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant
pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)

Applying the following property of exponentials numbers in (II):

e^{a}.e^{b}=e^{a+b}

Therefore e^{-28k} can be written as e^{-14k}.e^{-14k}

152=190-Ae^{-14k}.e^{14k}

Replacing (I) in the previous equation:

152=190-76e^{-14k}

Solving for k:

Subtracting 190 both sides, dividing by -76:

0.5=e^{-14k}

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76

Dividing by 0.5

A=152

7 0
3 years ago
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