New perimeter is 4040-2424= 1616
original, let x and y be length and width
2x + 2y = 4040
new
2x/2 +2y/3 = 1616
x +2y/3 = 1616
x = 1616 - 2y/3
sub this x value into first eq
2(1616-2y/3) + 2y = 4040
3232 - 4y/3 + 2 = 4040
put like terms together
-4y/3+2y=4040-3232
2y/3=808
multiply both sides by 3
2y = 2424
divide both sides by 2
y=1212
sub this y value into any eq. i chose eq 1
2x+2(1212)= 4040
2x = 4040 - 2(1212)
2x = 1616
x = 808
so the length is 808 and the width is 1212
Answer:
If we are working in a coordinate plane where the endpoints has the coordinates (x1,y1) and (x2,y2) then the midpoint coordinates is found by using the following formula:
midpoint=(x1+x22,y1+y22)
Step-by-step explanation:
This is the domain and range for exponential function
Answer:
The corresponding definite integral may be written as
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The answer of the above definite integral is
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Step-by-step explanation:
The given limit interval is
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![[a, b] = [-8, 6]](https://tex.z-dn.net/?f=%5Ba%2C%20b%5D%20%3D%20%20%5B-8%2C%206%5D)
The corresponding definite integral may be written as
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
Bonus:
The definite integral may be solved as

Therefore, the answer to the integral is
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