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Ierofanga [76]
2 years ago
13

Two consecutive numbers whose squares add up to 685

Mathematics
1 answer:
SVEN [57.7K]2 years ago
8 0

Question 790617: The sum of the squares of two consecutive integers is 685. Find the integers.

I tried solving this myself and got 324 and 325 which is not the right answer, I'm not very good with word problems. Thank you in advance!

(Scroll Down for Answer!)

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Found 2 solutions by rothauserc, stanbon:Answer by rothauserc(4570) About Me  (Show Source):

You can put this solution on YOUR website!

we are given the following

x^2 +(x+1)^2 = 685

x^2 +x^2+2x+1 = 685

2x^2 +2x = 684

x^2 +x = 342 and

x^2 +x -342 = 0

use quadratic equation to solve for x

x = (-1 + or - square root(1 - 4*1*(-342)) / 2

x = (-1 + or - 37) / 2

the problem does not state if our consecutive integers are positive or negative

x = 18 or -19

18^2 + 19^2 = 685

324 + 361 = 685

685 = 685

now

-19^2 + -18^2 = 685

361 + 325 = 685

so our integers are 18, 19, -19, -18

Answer by stanbon(75874) About Me  (Show Source):

You can put this solution on YOUR website!

The sum of the squares of two consecutive integers is 685. Find the integers.

--------

1st: x

2nd: x+1

------

Equation:

x^2 + (x+1)^2 = 685

x^2 + x^2 + 2x + 1 = 685

Step-by-step explanation:

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Area\ of\ material\ required\ for\ the\ first\ box=384\ inches^2\\Area\ of\ material\ required\ for\ the\ second\ box=486\ inches^2\\Area\ of\ material\ required\ for\ the\ first\ box=600\ inches^2\\Total\ Area\ of\ material\ required=1470\ inches^2

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We\ are\ given:\\Diameter\ of\ the\ first\ volleyball=8\ inches \\Diameter\ of\ the\ second\ volleyball=9\ inches\\Diameter\ of\ the\ third\ volleyball= 10\ inches.\\Hence,\\We\ know\ that,\\If\ the\ side\ of\ the\ cube\ box\ is\ s, it's\ Total\ Surface\ Area\ =No.\ of\\ faces\ in\ a\ regular\ polyhedron\ *Area\ of\ each\ face\ of\ the\ polyhedron=6*s^2=6s^2\\Hence,\\Lets\ apply\ this\ equation\ in\ finding\ the\ area\ of\ material\ required\ for\ the\\ three\ cases.\\

As\ the\ volleyball\ should\ wholly\ fit\ into\ the\ box,\ the\ diameter\ of\ the\\ volleyballs\ would\ be\ the\ side\ of\ the\ cube\ box.\\Hence,\\For\ the\ first\ volleyball,\\Diameter\ of\ the\ first\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ first\ volleyball=8\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ first\ box=6s^2=6*8*8=384\ inches^2

For\ the\ second\ volleyball,\\Diameter\ of\ the\ second\ volleyball=9\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ second\ volleyball=9\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ second\ box=6s^2=6*9*9=486\ inches^2

For\ the\ third\ volleyball,\\Diameter\ of\ the\ third\ volleyball=8\ inches\\Hence,\\Side\ of\ the\ cubical\ box\ for\ the\ third\ volleyball=10\ inches.\\Hence,\\The\ Total\ Surface\ Area\ of\ the\ third\ box=6s^2=6*10*10=600\ inches^2

Hence,\\If\ you\ are\ asked\ the\ Total\ Area\ to\ make\ all\ the\ boxes,\\ you\ just\ add\ them\ together.\\Hence,\\Total\ Area\ of\ Material\ required\ to\ make\ the\ three\ boxes=384+486+600=1470\ inches^2

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