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lana [24]
3 years ago
6

The length of a rectangle is 10 meters more than the width. If the perimeter is 184 meters, what are the length and the width?

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
3 0
Perimeter = (2 × Length) + (2 × Width)

If the length is dependent on the width (because the length is 10 more than the width), we can say that L = W + 10.

To find the answer to the Peri, we have to find out what the W is first.

184 = 2(W + 10) + 2W
184 = 2W + 20 + 2W
184 = 4W + 20
184 - 20 = 4W = 164
164 ÷ 4 = W
41 = W

We have found out width. Our length is W + 10, so L = 41 + 10 = 51

W = 41 and L = 51


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6 0
3 years ago
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10 points to the person who answers this whole thing and I will mark you as brainliest.
olga2289 [7]

Answer:

Step-by-step explanation:

Perimeter is found by adding together all the lengths of the sides. For us, that is x + (2x + 5) + (6x - 17) + (3x + 2). Now we will just combine like terms. We can also drop the parenthesis because they do nothing for us and mean nothing to the problem.

x + 2x + 5 + 6x - 17 + 3x + 2 becomes

12x - 10

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3 years ago
Suppose that a family wants to increase the size of the garden which is currently 16 feet long and 8 feet wide without changing
Cloud [144]

Answer:

See below

Step-by-step explanation:

<u>Sides of the garden</u>

  • 16 feet and 8 feet

<u>New dimensions of the garden</u>

  • 16 + 2x and 8 + x

<u>1. The area of the new garden</u>

  • (16 + 2x)(8 + x) =
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<u>2. The difference in area </u>

  • 2x^2 + 32x + 128 - 16*8 =
  • 2x^2 + 32x

<u>3. x= 2 effect on the area</u>

  • 2*2^2 + 32*2 =
  • 8 + 64 =
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<u>4. Area difference = 160</u>

  • 2x^2 + 32x = 160
  • x^2 + 16x - 80 = 0
  • Solving we get positive root of 4 ft

<u>5. Area difference = 4 times</u>

  • 2x^2 + 32x = 128*3
  • 2x^2 + 32x - 384 = 0
  • x^2 + 16x - 192= 0
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3 0
3 years ago
Help ASAP :) Brainly given + Thanks
horrorfan [7]

Answer:

80 sqcm

Step-by-step explanation:

area of the full rectangle is 96 sqcm (12x8) and then you subtract the area of the cutout (4x4)

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1 year ago
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10.(06.01 LC)
storchak [24]
Need more explanation
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