1. ## Help with formulas?

Would someone please explain to me how to solve these? I'm completely lost and can't even figure out one of the problems on my homework. I would greatly appreciate step-by-step instructions that are as non-confusing as possible.

A=115+1/2(p+s) solve for s=?

P=2w+2l solve for w=?

Also, this word problem:
Set up an equation for the given application problem.
The perimeter of standard - size copier papper is 99 cm. The width is 6.3 cm less than the length. Find the length and the width.

If you would like to just do one that's fine, I'm just in desperate need of some instructions.

2. Your first question is strange. I suppose its not looking for a concrete answer. in the first case, just isolate s to one side.
A-115-1/2p=1/2s
2A-230-p=s

Try this for the second one.

The perimeter of standard - size copier papper is 99 cm. The width is 6.3 cm less than the length. Find the length and the width.
Let x be length and y be width.

Perimeter = x+y+x+y = 2x+2y
99=2x+2y

y=x-6.3

Substituting, we get...

99=2x+2(x-6.3)
99=4x-12.6
111.6=4x
x=27.9cm
We know y is 6.3 less than this so,
y=21.6cm

3. Sol. 1)
$\ a= 115 + \frac{1}{2} (p+s)$
$\ a-15 = \frac{1}{2} (p+s) \$ or $\ \frac{1}{2} (p+s) =\ a-15$
$p+s =2(a-15)$
$s =2(a-15) -p$

4. sol. 2)
$P=2w+2l$
$P-2l=2w$
$2w=P-2l$
$w=\frac{1}{2}(P-2l)$

5. Ramiee, In sol 1, 15 should be replaced by 115.

6. oops

7. A = 115 + 1/2(p + s)

Don't wait; get rid of fraction RIGHT AWAY; multiply by 2:
2A = 230 + p + s

s = 2A - 230 - p

Easier, right?

### how to get an side of length and width if the with is 6.3less than length with a Perimeter of 99cm

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