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I'm kind of stuck on some problems on literal equations and formulas. Can someone help me? Also can u show me how u solved it?

Solve for x and indicate any restrictions on the values of the variables.
5ax-c=4c+ax
p(x-1)+7=r(x-4)

Solve each formula for the indicated variable. Indicate any restrictions on the values of the variables.
A=ptprt, for p
A=1/2h(b1+b2), for b1
E=i(R+r), for r

Solve for L
S=2(Lw+wh+Lh)

Thank u all soooooooo very much.

2006-08-20 08:49:09 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

5ax - c = 4c + ax
5ax - ax = 4c + c
x(5a - a) = 5c
x(4a) = 5c
x = (5c)/(4a)

"a" cannot be 0.

--------------------------------------

A = ptprt
A = p^2 * r * t^2
A/(rt^2) = p^2
p = sqrt(A/(rt^2))
p = (1/t)(sqrt(A/r))

"t" and "r" cannot be 0

-----------------------------------------

A = (1/2)h(b1 + b2)
2A = h(b1 + b2)
(2A)/h = b1 + b2
b1 = ((2A)/h) - b2

"h" cannot be 0

--------------------------------

E = i(R + r)
E/i = R + r
r = (E/i) - R

if 'i' is a varible, then "i" cannot equal 0

-------------------------

S = 2(Lw + wh + Lh)
S/2 = Lw + wh + Lh
S/2 = (Lw + Lh) + wh
(S/2) - wh = L(w + h)
((S/2) - wh)/(w + h) = L
L = ((S/2) - wh)/(w + h)

"w" and "h" cannot be opposite of each other.

2006-08-20 09:03:47 · answer #1 · answered by Sherman81 6 · 0 0

5ax-c=4c+ax
5ax-ax=4c+c
4ax=5c
x=5c/4a
a can't be zero

2006-08-20 16:25:21 · answer #2 · answered by Ethio /Habeshaw/ 3 · 0 0

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