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Ahat [919]
3 years ago
10

Hiram raises earthworms. In a square of compost 4 ft by 4 ft, he can have 1000 earthworms. How many earthworms can he have if hi

s square of compost has a side length that is 8 times longer?
Mathematics
2 answers:
charle [14.2K]3 years ago
6 0
64,000 I believe. What I did was find the area so 4*4 =16. 1000/16 = 62.5 earthworms per square foot. So if the side is 8 times longer then 8*4= 32 feet per side and to find the area of that it's 32*32 = 1024 feet squared. So 62.5 * 1024 =64,000
Anna11 [10]3 years ago
6 0

Answer:

The answer is 64000.

Step-by-step explanation:

Given is -

Hiram raises 1000 earthworms in a square of compost 4 ft by 4 ft.

When the square of compost has a side length that is 8 times longer, means the area will increase by a factor of 8\times8=64 times.

So, total earthworms will be = 1000\times64=64000

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Step-by-step explanation:

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andreev551 [17]

Recall the angle sum identity for cosine:

cos(<em>x</em> + <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) - sin(<em>x</em>) sin(<em>y</em>)

cos(<em>x</em> - <em>y</em>) = cos(<em>x</em>) cos(<em>y</em>) + sin(<em>x</em>) sin(<em>y</em>)

==>   sin(<em>x</em>) sin(<em>y</em>) = 1/2 (cos(<em>x</em> - <em>y</em>) - cos(<em>x</em> + <em>y</em>))

Then rewrite the equation as

sin(4<em>x</em>) sin(5<em>x</em>) + sin(4<em>x</em>) sin(3<em>x</em>) - sin(2<em>x</em>) sin(<em>x</em>) = 0

1/2 (cos(-<em>x</em>) - cos(9<em>x</em>)) + 1/2 (cos(<em>x</em>) - cos(7<em>x</em>)) - 1/2 (cos(<em>x</em>) - cos(3<em>x</em>)) = 0

1/2 (cos(9<em>x</em>) - cos(<em>x</em>)) + 1/2 (cos(7<em>x</em>) - cos(3<em>x</em>)) = 0

sin(5<em>x</em>) sin(-4<em>x</em>) + sin(5<em>x</em>) sin(-2<em>x</em>) = 0

-sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) (sin(4<em>x</em>) + sin(2<em>x</em>)) = 0

Recall the double angle identity for sine:

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Rewrite the equation again as

sin(5<em>x</em>) (2 sin(2<em>x</em>) cos(2<em>x</em>) + sin(2<em>x</em>)) = 0

sin(5<em>x</em>) sin(2<em>x</em>) (2 cos(2<em>x</em>) + 1) = 0

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… … … … … …    ==>   <em>x</em> = <em>π</em>/3 + <em>nπ</em>   <u>or</u>   <em>x</em> = -<em>π</em>/3 + <em>nπ</em>

<em />

(where <em>n</em> is any integer)

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