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11Alexandr11 [23.1K]
3 years ago
13

the area of a rectangular invitation card is 3/10 square foot. Given that the length of the card is 4/5 foot find its perimeter

Mathematics
1 answer:
zhenek [66]3 years ago
8 0
F=Foot

A=Area of the rectangular invitation card

L=Length of the rectangular invitation card

P=Perimeter

W=Width of the rectangular invitation card

------------------------

A=3/10f²

L=4/5f

-------------------------

A=L*W=3/10f²

So...

4/5f*W=3/10f²

So...

W=(3/10f²)/(4/5f)

So...

W=3/8f

--------------------------

Now:

P=2W+2L

=2(3/8f)+2(4/5f)

=3/4f+8/5f

=15/20f+32/20f

=47/20f

=(40/20+7/20)f

=(2+7/20)f

=2.35f
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Let's define the following variables first.

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C = number of tickets sold for children

From the question, we can say that or form the following equations:

1. A + C = 790 tickets

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The first equation can also be written as A = 790 - C. We can use this equation and replace "A" in the second equation.

7(790-C)+4c=4,390

From that, we can solve "C" by solving the equation formed above.

\begin{gathered} \text{Distribute 7 to the terms inside the parenthesis.} \\ 7(790)-7(C)+4C=4,390 \\ 5,530-7C+4C=4,390 \\ \text{Subtract 5,530 on both sides of the equation.} \\ -7C+4C=-1140 \\ -3C=-1140 \\ \text{Divide -3 on both sides.} \\ C=380 \end{gathered}

Therefore, 380 tickets for children were sold.

Since there are 790 tickets in total that are sold and 380 tickets for children were sold, we can say that 410 tickets for adult was sold.

\begin{gathered} A=790-C \\ A=790-380 \\ A=410 \end{gathered}

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