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3 years ago

8

owen has a collection of nickels and quarters worth $8.10. If the nickels were quarters and the quarters were nickels, the value

would be $17.70. Find the number of each coin.

1 answer:

prisoha [69]3 years ago

4 0

There are 67 nickels and 19 quarters.Im pretty sure

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Step-by-step explanation:

Since we have given that

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According to question, a cheese sandwich consists of 2 slices of bread and 3 slices of cheese.

So, we need to find the number of whole cheese sandwiches that can be prepared.

Number of sandwich containing only slice of bread is given by

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Number of sandwich containing only slice of cheese is given by

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As we know that each sandwich should contain both slice of bread and slice of cheese.

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Use properties to rewrite the given equation. Which equations have the same solution as 3/5x +2/3 + x = 1/2– 1/5x? Check all tha

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we have

$\frac{3}{5}x+ \frac{2}{3}+x=\frac{1}{2}-\frac{1}{5}x$

Combine like terms in both sides

$(\frac{3}{5}x+ x)+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

we know that

$(\frac{3}{5}x+ x)=(\frac{3}{5}x+ \frac{5}{5}x)=\frac{8}{5}x$

substitute in the expression above

$\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$ -----> equation A

Multiply equation A by $5*3*2=30$ both sides

$30*(\frac{8}{5}x+\frac{2}{3})=30*(\frac{1}{2}-\frac{1}{5}x)$

$48x+20=15-6x$ ---------> equation B

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$48x+6x=15-20$

$54x=-5$ ---------> equation C

Solve for x

$x=-\frac{5}{54} =-0.09$

We are going to proceed to verify each case to determine the solution.

<u>Case a)</u> $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

the case a) is equal to the equation A

so

the case a) have the same solution that the given equation

<u>Case b)</u> $18x+20+30x=15-6x$

Combine like terms in left side

$(18x+30x)+20=15-6x$

$(48x)+20=15-6x$

the case b) is equal to the equation B

so

the case b) have the same solution that the given equation

<u>Case c)</u> $18x+20+x=15-6x$

Combine like terms in left side

$(18x+x)+20=15-6x$

$(19x)+20=15-6x$

$19x+6x=15-20\\25x=-5\\x=-0.20$

$-0.20\neq -0.09$

therefore

the case c) not have the same solution that the given equation

<u>Case d)</u> $24x+30x=-5$

Combine like terms in left side

$54x=-5$

the case d) is equal to the equation C

so

the case d) have the same solution that the given equation

<u>Case e)</u> $12x+30x=-5$

Combine like terms in left side

$42x=-5$

$x=-5/42=-0.12$

$-0.12\neq -0.09$

therefore

the case e) not have the same solution that the given equation

therefore

<u>the answer is</u>

case a) $\frac{8}{5}x+\frac{2}{3}=\frac{1}{2}-\frac{1}{5}x$

case b) $18x+20+30x=15-6x$

case d) $24x+30x=-5$

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