(80 squared) + (3rd side squared) = (104.5 squared)
3rd side = 67.2328
2007-04-26 06:48:17
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answer #1
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answered by Rank Roo 4
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I love math!
Use the Pythagorean theorem
a^2 + b^2 = c^2
where a and b are the sides of a right triangle and c is the hypotenuse.
a^2 + 80^2 = 104.5^2
a^2 = 4520.25
a = 67.23
2007-04-26 06:53:41
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answer #2
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answered by Robert L 7
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Use your guy or woman math. i'm not yer consumate fool. i did not do the strolling. Now U declare to have walked 18 mi in 240 min. it somewhat is an well-known of four.5mph. this is somewhat speedier than the final human strolling velocity of three.2 mph. Are You a robotic or a marathon runner or a liar? i think U be the liar and a clever *** attempting to fool those that are too dumb to attain that eighty min=a million hr 20 min and does not suspect that 4.5 mph is a lot speedier than the final walk. So evaluate this a gotcha!
2016-12-10 12:08:44
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answer #3
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answered by cosner 4
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use a^2 + b^2 = c^2
a= 80 *80= 6400
b= ?
c= 104.5*104.5= 10920.25
now, take 10920.25-6400=4520.25
b=the square root of 4520.25= 67.2328= 67
2007-04-26 06:48:54
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answer #4
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answered by Anonymous
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A²+B²=C²
80²+B²=104.5²
6400+B²=10920.25
B²=10920.25-6400
Do the rest yourself.
2007-04-26 06:49:01
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answer #5
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answered by Barkley Hound 7
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Its very simple
104.5^2 = 80^2 + (3rd)^2
(3rd)^2 = 105^2 - 80^2
(3rd)^2 = 4625
take the square root of both sides
3rd = (4625)^.5
3rd = 68.00735
2007-04-26 06:51:10
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answer #6
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answered by ? 3
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The other side = sqrt. (104.5^2 - 80^2) = sqrt. 4520.25 = 67.23..
2007-04-26 06:48:29
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answer #7
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answered by Swamy 7
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A squared + B squared (the two legs) = C Squared, Hypotinuse is C (NOT C SQUARED) So C Squared - A Squared (one leg) is B Squared. Find the route and that's your answer.
2007-04-26 06:50:30
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answer #8
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answered by Whisper from the Myst 2
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Hyp. = sq. root (leg 1^2 + leg 2 ^2)
so -
Leg 2 = sq. root (Hyp.^2 - Leg 1 ^2)
I don't have a calculator on me, but you get the idea.
2007-04-26 06:50:26
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answer #9
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answered by peachfuzz 3
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10. 1324
2007-04-26 06:47:18
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answer #10
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answered by junkie 2
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