I think it should be (x+60)(x+40)=4800
So,
x^2+100x+2400=4800
x^2+100x-2400=0
(x+120)(x-20)=0
x=20..
so, it will be 60 meters wide and 80 meters long
2007-01-14 18:09:35
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answer #1
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answered by andru 2
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To multiply an area by 2 you multiply length and width by √2 = 1.414. Check it:
l • √2 • w • √2 = 2lw = 2A.
So each side should be extended by 41.4%. Ah, but that would not be equal amounts, would it. So,
(60+x)(40+x) = 2(2400) = 4800
x² + 100x - 2400 = 0
(x + 120)(x - 20) = 0
discard negative solution, x = 20.
Check: (40+20)(60+20) = 60(80) = 4800.
Why'd you use 2x? Confusing area and perimeter?
2007-01-14 18:14:16
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answer #2
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answered by Philo 7
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isn't the equation (x+60)(x+40)=4800?
in that case, i would just guess and check, though i think there's a better way.
if x=30, 90*70=4800. No.
if x=25, 85*65=4800. No.
if x=20, 80*60=4800. Yes!
Each side would be extended 20 meters.
if you didn't want the answer (just help), then i can only say, oops. Sorry.
2007-01-14 18:15:39
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answer #3
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answered by blah 3
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well.... you need to double the area. Knowing what the original area is might be a good first step.
A = Length * Width
so
A = 60 * 40 = 2400.
We want 2A, 2A = 4800.
We want to get to 2A by extending each side by an equal amount this gives: 2A = (L+x) * (W+x) or
2A = L*W+L*x+W*x+x^2
Now, solve for x.
2A = 2400+60x+40x+x^2
4800 = 2400 + 100x + x2
2400 = x^2+100x
0= x^2+100x - 2400
Now factor 2400. 120*20 works best and the equation factors as
0 = (x+120)*(x-20)
For this to zero out our two choices for x are x= -120 and 20. Since we can't add negative length our choice for x is 20 and we're done.
2007-01-14 18:24:11
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answer #4
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answered by anecdoteman1 2
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40*60=2400
(40+x) * (60+x)=4800
* = multiplies
If u want u can reverse the #s to say (x+40)(x+60)=4800
2007-01-14 18:06:18
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answer #5
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answered by Isabela 5
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the first ONE Label the first consecutive huge style x Label the 2d one x+a million sq. of the first = x^2 decreased by using 25 = x^2 - 25 Equals 3 circumstances the 2d: x^2 - 25 = 3(x +a million)
2016-10-31 03:29:00
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answer #6
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answered by ? 4
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I would go with Isabella. However, to check your answer, "FOIL" out your problem and solve with the quadratic formula. If you would like more help, please email me. I have no trouble at all.
2007-01-14 18:13:36
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answer #7
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answered by poetunknown2000 1
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