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3 years ago

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In a collection of nickels, dimes, andquarters the value of the collection is$12.85. If the number of nickels is 6 lessthan twic

e the number of dimes and thenumber of quarters is 5 more than twicethe number of dimes, how many quartersare there?

1 answer:

Sergio [31]3 years ago

7 0

30 quarters because you times 6 by 5 and that equals 30 than you divine that by 4 and that 7.5

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Step-by-step explanation:

To solve, we will just use some algebra

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We can confirm this by plugging it back into the original formula.

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For the problem 1/5g- 1/10- g + 1 3/10g -1/10, Tyson created an equivalent expression using the following steps. 1/5g+-1g+1 3/10

professor190 [17]

The true statements are:

Tyson's expression is not equivalent to the original expression
The equivalent expression is: $\frac{1}{2}g- \frac 1{5}$

<h3>What are equivalent expressions?</h3>

Equivalent expressions are expressions that have equal values

The original expression is given as:

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10}$

Collect like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac 15g - g + 1 \frac{3}{10}g- \frac 1{10} -\frac{1}{10}$

Evaluate the like terms

$\frac 15g- \frac 1{10}- g + 1 \frac{3}{10}g -\frac{1}{10} = \frac{1}{2}g- \frac 1{5}$

Tyler's equivalent expression is given as:

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10}$

Collect like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = \frac15g-g+ 1 \frac3{10}g-\frac 45g-\frac 1{10}+1 \frac{1}{10}$

Evaluate the like terms

$\frac15g-g+ 1 \frac3{10}g-\frac 1{10}-\frac 45g+1 \frac{1}{10} = -\frac 3{10}g$

The simplified expressions of the original expression, and Tyson's equivalent expressions are not equal.

Hence, Tyson's expression is not equivalent to the original expression

Read more about equivalent expressions at:

brainly.com/question/9603710

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2 years ago