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Maksim231197 [3]
4 years ago
10

52-8(ɳ-1) รѳʆѵɛ tɦiร Բѳʀ ɱɛ pʆɛɑรɛ

Mathematics
2 answers:
svetlana [45]4 years ago
8 0
Let's simplify step-by-step.

<span>52−<span>8<span>(<span>n−1</span>)

</span></span></span>Distribute:

<span>=<span><span>52+<span><span>(<span>−8</span>)</span><span>(n)</span></span></span>+<span><span>(<span>−8</span>)</span><span>(<span>−1</span>)

</span></span></span></span><span>=<span><span><span>52+</span>−<span>8n</span></span>+8

</span></span>Combine Like Terms:

<span>=<span><span>52+<span>−<span>8n</span></span></span>+8

</span></span><span>=<span><span>(<span>−<span>8n</span></span>)</span>+<span>(<span>52+8</span>)

</span></span></span><span>=<span><span>−<span>8n</span></span>+60

</span></span>Answer:

<span>=<span><span>−<span>8n</span></span>+<span>60</span></span></span>
Cloud [144]4 years ago
8 0
52-8(n-1)=52-8n+8=60-8n

Answer: 52-8(n-1)=60-8n
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Lizzie has 30 coins that total $4.80. All of her coins are dimes, D, and quarters, Q. Algebraically determine how many of each t
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<em><u>Solution:</u></em>

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We know that,

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Given that LIzzie has 30 coins

number of dimes + number of quarters = 30

d + q = 30 ---- eqn 1

Also given that the coins total $ 4.80

number of dimes x  value of 1 dime + number of quarters x  value of 1 quarter = 4.80

d \times 0.10 + q \times 0.25 = 4.80

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Let us solve eqn 1 and eqn 2

From eqn 1,

d = 30 - q ---- eqn 3

Substitute eqn 3 in eqn 2

0.1(30 - q) + 0.25q = 4.8

3 - 0.1q + 0.25q = 4.8

0.15q = 1.8

<h3>q = 12</h3>

From eqn 3,

d = 30 - 12

<h3>d = 18</h3>

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See below.

Step-by-step explanation:

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Step-by-step explanation:

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