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Alla [95]
1 year ago
10

You decide you need a new computer. the cost of the computer is $1128. however, the store also offers a rent to own option which

will cost $52 per week for 24 weeks. how much more will the rent to own option cost after you have made all of the payments?
Mathematics
1 answer:
Mazyrski [523]1 year ago
4 0

Answer:

$120

Step-by-step explanation:

What you have to do first is multiply the rent per week by the number of weeks. Then subtract the number of weeks by the cost of the computer.

1248 - 1128 = 120

Hope that helps!!! (:

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7 0
3 years ago
X - 2y = 3<br> 5x + 3y = 2<br> The lines whose equations are shown intersect at which point?
Lady_Fox [76]

The point of intersection of the two lines is at (1,-1)

<h3>System of equation</h3>

The given system of expression is shown below

x - 2y = 3

5x + 3y = 2

The solution to the system of equation is the point of intersection

From equation 1

x = 3 + 2y

Substitute into 2

5(3+2y) + 3y = 2

15 +10y + 3y = 2

13y = -13

y = -1

Substitute y = -1 into 3

x = 3 + 2y

x = 3+(-2)

x = 1

Hence the point of intersection of the two lines is at (1,-1)

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8 0
2 years ago
The simple interest on a certain sum of money for 2 years at 5% per annum is Rs 320. What will be the compound interest on the s
Brilliant_brown [7]

Answer:

Rs 328

Step-by-step explanation:

Find the <u>principal</u> amount invested.

<u>Simple Interest Formula</u>

I = Prt

where:

  • I = interest earned
  • P = principal
  • r = interest rate (in decimal form)
  • t = time (in years)

Given:

  • I = Rs 320
  • r = 5% = 0.05
  • t = 2 years

Substitute the given values into the formula and solve for P:

⇒ 320 = P(0.05)(2)

⇒ 320 = P(0.1)

⇒ P = 3200

<u>Compound Interest Formula</u>

\large \text{$ \sf I=P\left(1+\frac{r}{n}\right)^{nt} -P$}

where:

  • I = interest earned
  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Given:

  • P = 3200
  • r = 5% = 0.05
  • n = 1 (annually)
  • t = 2 years

Substitute the given values into the formula and solve for I:

\implies \sf I=3200\left(1+\frac{0.05}{1}\right)^{2} -3200

\implies \sf I=3200\left(1.05\right)^{2} -3200

\implies \sf I=3200\left(1.1025\right) -3200

\implies \sf I=3528-3200

\implies \sf I=328

Therefore, the compound interest on the same sum for the same time at the same rate is Rs 328.

7 0
2 years ago
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