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PIT_PIT [208]
3 years ago
5

Chris rode his bike 4 miles west and then 3 miles south. What is the

Mathematics
1 answer:
melamori03 [73]3 years ago
7 0

Answer:

5 miles

Step-by-step explanation:

This is because if you draw a line going west and label it as 4 miles, then draw another line going south and label it 3 miles, you form a right angle, right? So, the fastest way to go back would be taking a diagonal-like route back to the beginning. Does this shape look familiar to you? Right! It's a right triangle. And do the numbers 3, 4, and 5 sound familiar to you? Yup, it's a Pythagorean triplet. If the legs of a right triangle are 3 and 4, the hypotenuse is 5. Why am I saying this? Because, in the drawing you just did, the 4 miles west and 3 miles south were the legs, and the diagonal path back to the beginning is the hypotenuse. Get it now?

Happy to help! :)

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PLEASE HELP.<br> What is the value of X in the photo below? (please show work)
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A rectangular storage container with an open top is to have a volume of 10 m3. The length of this base is twice the width. Mater
Lelechka [254]
L=2W, V=LWH using L=2W in the Volume equation we get:

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10=2W^2H  now we can solve this for H

H=5/W^2 and L=2W  we'll need these later :)

C=20LW+12*2LH+12*2WH

C=20LW+24LH+24WH  using our H and L found earlier...

C=20(2W^2)+24(2W*5/W^2)+24(W*5/W^2)

C=40W^2+240/W+120/W  making a common denominator...

C=(40W^3+240+120)/W

C=(40W^3+360)/W

dC/dW=(120W^3-40W^3-360)/W^2

dC/dW=(80W^3-360)/W^2

d2C/dW2=(240W^4-160W^4+720W)/W^4

d2C/dW2=(80W^3+720)/W^3

Since d2C/dW2 is positive for all possible values of W (as W>0), when dC/dW=0, C(W) will be at an absolute minimum value...

dC/dW=0 only when 80W^3-360=0

80W^3=360

W^3=45

So our minimum cost is:

C(45^(1/3))=(40W^3+360)/45^(1/3)

C(45^(1/3))=(40*45+360)/45^(1/3)

C(45^(1/3))=2160/45^(1/3)

C≈$607.27  (to the nearest cent)


8 0
4 years ago
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