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Topic: I need help with an algebra word problem i need to solve for x,y i am having trouble setting up the problem?**Question:**
the problem is: the perimeter of a badminton court is 128ft. after a game of badminton a player's coach estimates that the athlete has run a total of 444ft, which is equivalent to six time the courts length plus nine times it width. What are the dimensions of a standard badminton court
the answer is 44 by 20ft the problem im having is setting up the problem

July 19, 2019 / By Shulamite

L=length, W=Width 2L+2W=128 6L+9W=444 multiply the first equation by 3... 6L+6W=384 You then have two equations, both with "6L" so you can subtract the second equation from the first 6L+9W=444 6L+6W=384 {6L-6L=0} {9W-6W=3W} {444-384=60} so.... 3W=60 W=20 substitute back into the first equation 2L+2(20)=128 2L+40=128 2L=88 L=44 checking 6L+9W=444 6(44)+9(20)= 264+180=444 good luck

👍 232 | 👎 2

Did you like the answer? From the question we know: L=3s-80 m=s+15 The sum of a triangle's angles is 180 so: L+m+s=180 Let's find the smallest first: Since L=3s-80 and m=s+15, then (3s-80)+(s+15)+s=180 And simplify: 3s+s+s-65=180 5s=245 s=49 Now the large angle: Plug in 49 for s L=3(49)-80 L=67 And the middle: m=49+15 m=64

First you must set up what your equation needs to be, in this case... x= smallest angle x+15= 2nd smallest angle 3x-80= largest angle Then you add x+x+15+3x-80 to get... 5x-65 Then you need to set 5x-65 against 180 because there are 180 degrees in a triangle. The equation will now look like 5x-65=180 Add 65 to both sides to get.. 5x=245 Then divide by 5 on both sides to get... x=49 Then plug x back into your angles you had at the beginning to get... Smallest angle: 49 Middle angle: 64 Large angle: 67 If you want you can add all these up to check your work and get 180. Hope this helps.

Lenght is L; Width is W 2w + 2l = 128 6l + 9w = 444 You can use linear combination or substitution. Using substitution: 2w = 128 - 2l w = 64 -l 6l + 9(64-l) = 444 6l + 576 - 9l = 444 - 3l = - 132 l = 44 w = 64 - 44 w = 20

👍 100 | 👎 1

If you say x=side of square that must be cut out. Volume is LWH H = x L = 14 -2x W = 11 - 2x Thefefore x(14-2x)(11-2x) = 80 Solve for x

The third answer is correct: L=14-2h W=11-2h 80=LWh Substituting in to eliminate L and W we get h^3-12.5h^2+38.5h-20=0 h=.65 is one solution h=3.85 is another

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