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zalisa [80]
3 years ago
5

A rectangle and a square have the same area. The width of the rectangle is 2 in less than the side of the square and the length

of the rectangle is 3 in less than twice the side of the square. What are the dimensions of the rectangle?
Mathematics
1 answer:
lina2011 [118]3 years ago
5 0
The area of a rectangle is A=LW, the area of a square is A=S^2.

W=S-2 and L=2S-3

And we are told that the areas of each figure are the same.

S^2=LW, using L and W found above we have:

S^2=(2S-3)(S-2)  perform indicated multiplication on right side

S^2=2S^2-4S-3S+6  combine like terms on right side

S^2=2S^2-7S+6  subtract S^2 from both sides

S^2-7S+6=0  factor:

S^2-S-6S+6=0

S(S-1)-6(S-1)=0

(S-6)(S-1)=0, since W=S-2, and W>0, S>2 so:

S=6 is the only valid value for S.  Now we can find the dimensions of the rectangle...

W=S-2 and L=2S-3  given that S=6 in

W=4 in and L=9 in

So the width of the rectangle is 4 inches and the length of the rectangle is 9 inches.


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Pls help for brainliest answer
lina2011 [118]

Answer:

A. x = 24

B. x = 20

Step-by-step explanation:

<u>→A. Add 2 to both sides:</u>

<u />\frac{x}{3} -2=6

\frac{x}{3} =8

→The fraction \frac{x}{3} is really just \frac{1}{3}x:

<u>→Multiply 3 to both sides:</u>

x = 24

<u>→B. Subtract 1 from both sides:</u>

<u />\frac{x}{5} +1=5

\frac{x}{5} =4

The fraction \frac{x}5} is really just \frac{1}{5}x:

<u>→Multiply 5 to both sides:</u>

x = 20

8 0
3 years ago
8x + 4y = 2 solve for y
Ivanshal [37]
The quickest way to solve this is 

8(0)+4y=2 (since we are looking for y we are going to put 0 in the x intercept) with that all we have left is this

4y=2 next we divide 4 on both sides in order to get y by itself. 
2/4=0.5
y=0.5 or 1/2.
5 0
4 years ago
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Plz help me id rly appreciate it
natulia [17]

Answer:

\frac{n^{10} }{16m^{6} }

Step-by-step explanation:

Photomath

5 0
3 years ago
Westfalls is 7 miles south of edenville and Concord is 13 mile west of westfalls. What is the distance from edenville to Concord
Lelu [443]

Answer:

D_x\approx14.8mile

Step-by-step explanation:

From the question we are told that

Distance between Edenville and Westfalls D_1=7

Distance between  Westfalls  and Concord D_2=13

Generally the equation for Pythagoras is mathematically given by

D_x^2=D_1^2+D_2^2

D_x^2=7^2+13^2

D_x=\sqrt{ 7^2+13^2}

D_x=14.76482306mile

The distance b/w Edenville to Concord is

D_x\approx14.8mile

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3 years ago
The difference of m and 7 increase by 15
uranmaximum [27]
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