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valina [46]
3 years ago
4

A designer enlarged both the length and the width of a rectangular carpet by 60 percent. The new carpet was too large so the des

igner was asked to reduce its length and its width by 25 percent. By what percent was the area of the final item greater than the area of the original?
Mathematics
1 answer:
Naily [24]3 years ago
3 0

Answer:

300%

Step-by-step explanation:

Let the length of the carpet be L.

Let the width of the carpet be W.

The original area of the carpet is:

A = L * W = LW

The length and width are enlarged by 60%.

The new length is:

L + (L * 60/100) = L + 0.6L = 1.6L

The new width is:

W + (W * 60/100) = W + 0.6W = 1.6W

The length and width are then reduced by 25%.

The new length is:

1.6L + (1.6L * 25/100) = 1.6L + 0.4L = 2L

The new width is:

1.6W + (1.6W * 25/100) = 1.6W + 0.4W = 2W

The new area will be:

A = 2L * 2W = 4LW

To find the percentage increase in the area, we subtract the original area from the new area and divide by the original area:

4LW - LW = 3LW

% increase is:

\frac{3LW}{LW}  * 100 = 300%

The area of the final item is 300% greater than the original item.

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Marizza181 [45]

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

In the given figure, the Angle (6v - 4)° forms vertical opposite angle pair with 50° , so their measures is equal to one another.

that is :

  • 6v - 4 = 50

  • 6v = 50 + 4

  • 6v = 54

  • v =  \dfrac{54}{6}

  • v = 9

therefore, value of v = 9°

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3 years ago
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Please someone help me
White raven [17]

Answer:

i think the answer is 36.......

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3 years ago
Please help me with the problem
mixas84 [53]
Let's see what we're working on here 

\frac{1}{2}  - 2(m) =?
            ? = 0
Simplify this → \frac{1}{2}

\frac{1}2y} - 2(m) = ?

\frac{2(m)(2)}{2}

\frac{1 - 4(m)}{2} = ?

\frac{1 - 4(m)}{2} = 2

-4(m) + 1 = 0
 Subtract ( - )the number 1 from each of your sides on this part

4(m) = 1 
Multiply( ×) the number -1 to your sides for this part of the equation 

Therefore, the value of m is \frac{1}{4}
m =  \frac{1}{4}


3 0
4 years ago
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abbie spent 5/8 of her money and saved the rest. if she spent 45,how much money did she have at first?
algol13

$72

<h3>Further explanation</h3>

<u>Given:</u>

Abbie spent ⁵/₈ of her money and saved the rest.

She spent $45.

<u>Question:</u>

How much money did she have at first?

<u>The Process:</u>

The meaning of "Abbie spent ⁵/₈ of her money" is that she has spent 5 of the 8 parts of the money she had at first.

Let us make a pattern of 8 units for the amount of money that Abbie had at first.

\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}\boxed{\cdot}

Since Abbie spent ⁵/₈ of her money, then we create a pattern of 5 units containing $9 each. Recall that \boxed{\$ 45 \div 5 = \$ 9}

\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9} = \$ 45

From the pattern above, we can see that \boxed{ \ 1 \ unit = \$ 9 \ }.

Let us calculate how much money Abbie had at first.

\boxed{ \ 1 \ unit = \$ 9 \ }

\boxed{ \ 8 \ unit = \ ? \ }

\boxed{ \ 8 \times \$ 9 = \$ 72 \ }

Thus, the money Abbie had at first is $72.

The complete pattern of Abbie's initial money is as follows:

\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9}\boxed{\$ 9} = \$ 72

- - - - - - - - - -

<u>Quick Steps:</u>

Let M as Abbie's initial money.

\boxed{ \ \frac{5}{8} \times M = \$ 45 \ }

Both sides are multiplied by ⁸/₅ so that M is the subject alone on the left side.

\boxed{ \ M = \frac{8}{5} \times \$ 45 \ }

We crossed out 5 and 45.

\boxed{ \ M = 8 \times \$ 9 \ }

\boxed{ \ M = \$ 72 \ }

Thus, the money Abbie had at first is $72.

<h3>Learn more</h3>
  1. How much did Ciro spend altogether for the concert ticket, hotel, and dinner? brainly.com/question/12507869
  2. How much does his cat weigh (Jesse's)? brainly.com/question/388037
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Keywords: Abbie, spent 5/8 of her money, saved the rest, $45, $72, how much money, she have at first, the pattern

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Answer:

<em>A) (-5,7)</em>

Step-by-step explanation:

<u>Functions and Relations</u>

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\boxed{A)\ (-5,7)}

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