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Algebra 13 Online
OpenStudy (anonymous):

with a radius of 3x. how much of the land will be left over after the garden is built ?

OpenStudy (anonymous):

hold on - i messed up thats not the whole thing

OpenStudy (anonymous):

Is there more to this question

OpenStudy (anonymous):

a community group is building a garden on a rectangular lot of land the lot measures 5x by 10x and the garden will be circular with a radius of 3x. how much of the land will be left over after the garden is built ?

OpenStudy (anonymous):

how much land is there to begin with? ^_^

OpenStudy (anonymous):

|dw:1383106993167:dw|

OpenStudy (anonymous):

Simple area problem.

OpenStudy (anonymous):

|dw:1383106975954:dw| diameter of 6x (radius of 3x) doesn't fit in the garden. Instructions said circular not oval.

OpenStudy (anonymous):

|dw:1383107103183:dw|

OpenStudy (anonymous):

A.)41 pi X^2 b.)x^2(50-9 pi) C.)x^2(50-3 pi) d.)41x^2 pi=3.14

OpenStudy (anonymous):

ehuman is right, it wont fit in a 5x by 10x garden...

OpenStudy (anonymous):

but, (5x*10x) - 3.14*9 = 40.1x^2

OpenStudy (anonymous):

Area of circle is \[\pi * r ^{2}\ = 9x ^{2}] Area of rectangle is length x width = 50x.

OpenStudy (anonymous):

9x^2 x pi = 50 x

OpenStudy (anonymous):

9x^2 x pi = 50x^2. sorry

OpenStudy (anonymous):

how is the area of the 3x radius circle equal to the 5x by 10x rectangle?

OpenStudy (anonymous):

so it would be this right ?\[x^2(50-9\pi)\]

OpenStudy (anonymous):

Move equation to one side\[9x ^{2}\pi - 50x ^{2} = 0\] Factor out x^2 and you get \[x ^{2}(-50 + 9\pi)\]

OpenStudy (anonymous):

Yeah sorry i messed that up but it would be b.

OpenStudy (anonymous):

That equation should be switched\[50x ^{2} - 9x ^{2}\pi = x^{2}(50 - 9\pi)\]

OpenStudy (anonymous):

oh okay , ;) THANKS FOR ALL OF YA'LL HELP :)

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