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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
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mandy wrote:
. :? Hello
need help thanks
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?
(A) w^2+pw+k=0
(B) w^2-pw+2k=0
(C) 2 w^2+pw+2k=0
(D) 2 w^2-pw-2k=0
(E) 2w^2-pw+2k=0



C :

K = [(P-2W)/2] *W
thus 2k = PW - 2W^2
= 2w^2 - PW + 2k = 0
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
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mandy wrote:
. :? Hello
need help thanks
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?
(A) w^2+pw+k=0
(B) w^2-pw+2k=0
(C) 2 w^2+pw+2k=0
(D) 2 w^2-pw-2k=0
(E) 2w^2-pw+2k=0


Another way to solve the problem is to choose a rectangle of your own.

I chose one , in which each side was 1 unit.
hence w =1 , p =4 and k = 1.

Only E satisfied these values.

HMTG.
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
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mandy wrote:
. :? Hello
need help thanks
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?
(A) w^2+pw+k=0
(B) w^2-pw+2k=0
(C) 2 w^2+pw+2k=0
(D) 2 w^2-pw-2k=0
(E) 2w^2-pw+2k=0


Let l be the length of the board, then p = 2l + 2w, k = w*l
By plugging, we can find that the answer is E:
\(2w^2 - w(2l + 2w) + 2w*l = 0\), thus
2w - 2l - 2w + 2l = 0
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
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mandy wrote:
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?

(A) w^2 + pw + k = 0
(B) w^2 - pw + 2k = 0
(C) 2w^2 + pw + 2k = 0
(D) 2w^2 - pw - 2k = 0
(E) 2w^2 - pw + 2k = 0


Similar question to practice: an-equilateral-triangle-is-inscribed-in-a-circle-if-the-130556.html
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
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mandy wrote:
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?

(A) w^2 + pw + k = 0
(B) w^2 - pw + 2k = 0
(C) 2w^2 + pw + 2k = 0
(D) 2w^2 - pw - 2k = 0
(E) 2w^2 - pw + 2k = 0


lets say length is X.

area K = WX, hence X= K/W
perimeter P = 2W+2X

P = 2W + 2K/W
PW = 2W^2 + 2K

2w^2 - pw + 2k = 0 -- E
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
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Let x = length
p=2w+2x ===> x=1/2p-w

k=wx ===> x=k/w

Substituting:

k/w=1/2p-w
k=1/2pw-w^2
2w^2-pw+2k=0

Answer E.
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
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mandy wrote:
A rectangular circuit board is designed to have width w inches, perimeter p inches, and area k square inches. Which of the following equations must be true?

(A) w^2 + pw + k = 0
(B) w^2 - pw + 2k = 0
(C) 2w^2 + pw + 2k = 0
(D) 2w^2 - pw - 2k = 0
(E) 2w^2 - pw + 2k = 0


We can let n = the length of the rectangle and create the following equation:

2w + 2n = p

2n = p - 2w

n = (p - 2w)/2

Since area = n x w:

k = (n)(w)

k = [(p - 2w)/2]w

2k = pw - 2w^2

2w^2 - pw + 2k = 0

Answer: E
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
If you're just terrible at geometry, just remember that:

P=2(l+w)
A=wl=k inches^2

Just with this, you know there must be 2w in the equation, which eliminates A & B, which will then lead you to logic: C is too big, D is too small, E is just balanced: Bang, it's found in 45 seconds :')
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
Weird question. I didn't even really understand what it was asking initially...took me 4 minutes.

I did it algebraically.

A = lw
P = 2l + 2w

We can quickly eliminate A and C because those will always be positive.

Plug in the equations above for the remainder.

E is the answer.
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
Another way to solve it, allthough already explaind Plugin values,
say w= 2
so P= 8
and K= 4
Lets put thesde values in options, only E satisfies.
As pointed out by expert choice A&C can be directly discarded as sum of positive values never be zero.

Amar P
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
Start with defining a rectangle that has width of W (mentioned in the question) and length of L (NOT mentioned in the question).

The question stem provides that the AREA of the rectangle is equal to K, so with our variables above we can say that LW = K. Since L is NOT mentioned in the question we can set L to be in terms of variables that ARE mentioned in the question stem which is K/W, so now we have a rectangle with length of K/W and width of W.

Finally the question stem states that the perimeter is equal to P. So 2K/W + 2W = P, which means that E is the correct answer.
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Re: A rectangular circuit board is designed to have width w inches, perime [#permalink]
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