The width of a rectangle is 12 units. Can the perimeter x of the rectangle can be 60 units when its length y is 18 units?
A:No, the rectangle cannot have x = 60 and y = 18 because x + y is less than 60
B:No, the rectangle cannot have x = 60 and y = 18 because x + y is less than 24
C:Yes, the rectangle can have x = 60 and y = 18 because x = 12 + 2y
D:Yes, the rectangle can have x = 60 and y = 18 because x = 24 + 2y
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OpenStudy (anonymous):
@perl @dan815 @wio
OpenStudy (anonymous):
@mathmath333
OpenStudy (anonymous):
@Hero
OpenStudy (anonymous):
@hughfuve
OpenStudy (anonymous):
@Hero @mathmath333 @perl @dan815 @wio
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OpenStudy (anonymous):
@Mehek14 can you help me please
Mehek (mehek14):
|dw:1434755438008:dw|
Mehek (mehek14):
so add up 18+18+12+12 to find the perimeter
OpenStudy (anonymous):
60
Mehek (mehek14):
correct so the perimeter can be 60
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Mehek (mehek14):
that means that you can eliminate A and B
OpenStudy (anonymous):
ok
Mehek (mehek14):
now the formula for perimeter is P = 2w + 2l
w = width
l = length
Mehek (mehek14):
in your case length is y
so P = 2w + 2y
Mehek (mehek14):
you know that the width is 12
so P = 2(12) + 2y
2*12 = 24
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Mehek (mehek14):
so your formula would be
x = 24 + 2y
OpenStudy (anonymous):
So D
Mehek (mehek14):
yes
OpenStudy (anonymous):
Can you help with one more
Mehek (mehek14):
sure
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OpenStudy (anonymous):
ill open a new question so u can get another medal