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USPshnik [31]
2 years ago
14

If a rectangle's length is 7 more than 4 times the width and the perimeter is 54 what are the dimensions of the rectangle?

Mathematics
1 answer:
Rus_ich [418]2 years ago
4 0
Let w = width of the rectangle 
<span>4w+7 = length of the rectangle </span>

<span>perimeter = 2 lengths + 2 widths </span>

<span>2w + 2(4w+7) = 54 </span>
<span>2w + 8w + 14 = 54 </span>
<span>10w = 40 </span>
<span>w = 4 </span>
<span>4w+7 = 23 </span>

<span>dimensions of rectangle: </span>
<span>width = 4 </span>
<span>length = 23 hope this helps</span>
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Answer:

x=2(twice)

y=0(twice)

Step-by-step explanation:

This question can be solved using substitution method

So let's solve

y=x+2....(1)

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Substitute (1) into(2)

X+2=x2+5x+6

Collect like terms

X2+5x-x+6-2=0

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X2+2x+2x+4=0

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3 years ago
How many times can 8 go into 9
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Answer:

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Step-by-step explanation:

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8 0
3 years ago
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kenny6666 [7]

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Step-by-step explanation:

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no

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7 0
2 years ago
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Please write answers to the questions that are in the image.
marishachu [46]

Answer:

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His mistake for a is that he didn't find the common denominator and that he just added the fraction as it is. So the corrected version would be 14/15.

To find part B you also need to find the least common denominator.

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Step-by-step explanation:

8 0
3 years ago
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