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liubo4ka [24]
3 years ago
9

Jason went to the post office and bought both 41-cent stamps and 26-cent postcards and spent $10.28. The number of stamps was 4

more than twice the number of postcards. How many of each did he buy?
Mathematics
1 answer:
Charra [1.4K]3 years ago
8 0
<h3>Jason bought 20 stamps of $0.41 each and 8 postcards of $0.26 each.</h3>

<em><u>Solution:</u></em>

Let stamps be s and postcards be p

Given that,

The number of stamps was 4 more than twice the number of postcards

s = 4 + 2p -------- eqn 1

Jason bought both 41-cent stamps and 26-cent postcards and spent $10.28

41 cent = $ 0.41

26 cent = $ 0.26

Therefore,

s \times 0.41 + p \times 0.26 = 10.28

0.41s + 0.26p = 10.28 --------- eqn 2

Substitute eqn 1 in eqn 2

0.41(4 + 2p) + 0.26p = 10.28

1.64 + 0.82p + 0.26p = 10.28

1.08p = 10.28 - 1.64

1.08p = 8.64

Divide both sides by 1.08

p = 8

Substitute p = 8 in eqn 1

s = 4 + 2(8)

s = 4 + 16

s = 20

Thus Jason bought 20 stamps and 8 post cards

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Answer:

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Step-by-step explanation:

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we know that

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step 1

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<u><em>Find the distance BC</em></u>

substitute the values

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<u><em>Find the distance CD</em></u>

substitute the values

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CD=5\ units

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substitute the values

d=\sqrt{(-7+7)^{2}+(0-4)^{2}}

d=\sqrt{(0)^{2}+(-4)^{2}}

AD=4\ units

Compare the length sides

AB=CD

BC=AD

therefore

Opposite sides are congruent

step 2

Find the length of the diagonals

<u><em>Find the distance AC</em></u>

substitute the values

d=\sqrt{(-2+7)^{2}+(0-4)^{2}}

d=\sqrt{(5)^{2}+(-4)^{2}}

AC=\sqrt{41}\ units

<u><em>Find the distance BD</em></u>

substitute the values

d=\sqrt{(-7+2)^{2}+(0-4)^{2}}

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BD=\sqrt{41}\ units

Compare the length of the diagonals

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therefore

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