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Mathematics 43 Online
OpenStudy (anonymous):

The area of a rectangle is 84cm^2. The perimeter is 40cm find the dimensions.

OpenStudy (saifoo.khan):

xy = 84 --- 1 2y+2x = 40 ---2

OpenStudy (saifoo.khan):

Solve by substituion!

myininaya (myininaya):

Area of rectangle=A=xy=84 P=2x+2y=40

myininaya (myininaya):

xy=84 2x+2y=40=> x+y=20

OpenStudy (anonymous):

do it the cool way! solve this quadratic equation: \[p^2-20p+84 = 0\]

OpenStudy (anonymous):

so 5 and 15

myininaya (myininaya):

lol the cool way

OpenStudy (anonymous):

@CabRamos youre close, but 15*5 doesnt equal 84, so thats not the correct answer >.<

myininaya (myininaya):

y(x+y)=y(20) xy+y^2=20y y^2+xy-20y=0 y^2+84-20y=0 y^2-20y+84=0 (y-6)(y-14)=0

OpenStudy (anonymous):

6 and 14 then?

OpenStudy (anonymous):

nice :)

OpenStudy (anonymous):

Thanks ha

myininaya (myininaya):

gj!

OpenStudy (anonymous):

could it also be 16 and 4?

OpenStudy (anonymous):

yes that is correct too

OpenStudy (anonymous):

cause both of those are options on my worksheet

myininaya (myininaya):

16 and 4? 16(4) is not 84

OpenStudy (anonymous):

whoops >.< 14 and 6. i got confused >.<

OpenStudy (anonymous):

i thought he just switched the order >.< my bad my bad.

myininaya (myininaya):

if y=6, then x=14 if y=14, then x=6

OpenStudy (anonymous):

joe did u ever figure out the pool question?

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