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Vlad1618 [11]
3 years ago
14

Carter took 7 coins out of his pocket. He did not have any pennies.He had more than $0.95 and less than $1.13 what could his coi

ns have been ?Prove your anser
Mathematics
1 answer:
RUDIKE [14]3 years ago
4 0

Answer:

Carter's coins could be;

3 of 5¢, 2 of 10¢, 1 of 25¢, and 1 of 50¢

Step-by-step explanation:

The available coin denominations are;

1¢ = 1 penny, 5¢, 10¢, 25¢, 50¢, and $1.00

Since the coins are more than $0.95, we can have;

5¢ × 3

10¢× 2  

25¢× 1

50¢

Total = $1.00

That is his coins could have been 3 × 5¢, 2 × 10¢, 1 × 25¢, and 1 × 50¢, to make a total of 7 coins with an amount value of $1.00.

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Answer:

Step-by-step explanation:

To find the answer to Part A, we can turn this into a simple equation:

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135.2 = Answer to Part A.

Now, onto Part B. We can use <u>another</u> equation!

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Now, we must solve the given equation, first by taking 100.7 from each side:

192.7 - 2.3x = 100.7

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We are now left with: 92 - 2.3x = 0 - we must now add 2.3x to 2.3x and 0, giving us this:

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I hope this was really helpful! Tell me if you need any more help.

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A group of artists is selling small prints to raise money for charity. They sell p prints, and make a profit of f(p) on the sale
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Answer:

Statement A is true.

Step-by-step explanation:

A. If the group sells 15 prints, they will lose $85.

B. If the group sells 12 prints, they will lose $204.

C. If the group sells 35 prints, they will make $935.

D. If the group sells 28 prints, they will lose $136.

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Let's check the statements to see which one is true.

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Answer:

A and B

Step-by-step explanation:

Since these both only have one term and number/variable, making it a monomial.

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Answer:

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