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Nesterboy [21]
3 years ago
11

a rectangular lot is 80 meters wide and 115 meters long give the length and width of another rectangle that had the same perimet

er but a larger area
Mathematics
1 answer:
mart [117]3 years ago
7 0
1st...

80 * 115
P = 2 * 80 + 2 * 115
P = 160  + 230
P = 390 is the perimeter

A = 80 * 115 
A = 9, 200 is  the area.



I could only get the first part done for you maybe others are watching where they can help you with the second part of the question:::




Hope that helps!!!!


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Where is the options

Step-by-step explanation:

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Given a rectangle with an area of 20 square units, if the width is x units and the length is x + 1 units, what is the difference
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Answer: 1unit.

Step-by-step explanation:

Area of the rectangle

A = length × width

Since the area = 20 and the width is x and the length is x + 1.

We now substitute for the values in the above formula and solve for x

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20 = x( x + 1 ), we now open

20 = x² + x , then re arrange.

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Here, I am going to solve using factorization by grouping

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Therefore, the solution for x will be

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5 - 4 = 1unit.

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Jason ran 325 meters more than kim. kim ran 4.2 kilometers .how many meters did jason run?
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I know the answer: It is 329.2!!:)

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A number n plus 8 is greater than 11? What is the inequality and the solution?
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Ann is adding water to a swimming pool at a constant rate. The table below shows the amount of water in the pool after different
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Answer:

(a)  As time increases, the amount of water in the pool increases.

     11 gallons per minute

(b)  65 gallons

Step-by-step explanation:

From inspection of the table, we can see that <u>as time increases, the amount of water in the pool increases</u>.

We are told that Ann adds water at a constant rate.  Therefore, this can be modeled as a linear function.  

The rate at which the water is increasing is the <em>rate of change</em> (which is also the <em>slope </em>of a linear function).

Choose 2 ordered pairs from the table:

\textsf{let}\:(x_1,y_1)=(8, 153)

\textsf{let}\:(x_2,y_2)=(12,197)

Input these into the slope formula:

\textsf{slope}\:(m)=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{197-153}{12-8}=\dfrac{44}{4}=11

Therefore, the rate at which the water in the pool is increasing is:

<u>11 gallons per minute</u>

To find the amount of water that was already in the pool when Ann started adding water, we need to create a linear equation using the found slope and one of the ordered pairs with the point-slope formula:

y-y_1=m(x-x_1)

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\implies y-153=11x-88

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When Ann had added no water, x = 0.  Therefore,

y=11(0)+65

y=65

So there was <u>65 gallons</u> of water in the pool before Ann starting adding water.

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2 years ago
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