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Major math crisis here PLEASE HELP!!!!!!!!
PostPosted: Fri Apr 22, 2005 1:13 pm
by Artist4Jesus89
PostPosted: Fri Apr 22, 2005 1:26 pm
by Icarus
Is that A = (h/2) (b+c) or A= h(2(b+c)) ?
Whicheve it is, what you need to do is manipulate it until you have h by itself on one side. You use arithematic to slowly move everything to the other side of the equation.
Say you had
X = (25*Y)/Z +18
and you had to solve for Y. First, you would add (-18) to both sides:
X - 18 =(25*Y)/Z
(X-18)*Z = 25Y
[ (X-18)*Z ]/25 = Y
PostPosted: Fri Apr 22, 2005 1:33 pm
by Artist4Jesus89
thanks
PostPosted: Fri Apr 22, 2005 1:41 pm
by Icarus
Almost anything before writing a paper.
Which is to say, you're welcome.
PostPosted: Fri Apr 22, 2005 1:53 pm
by Syaoran
Dose this help you out
A=(h/2)(b+c)
A/(b+c)=h/2
h=2(A/(b+c))
I Manipulated it untill I solved for h. when it says solve for h or a or what ever it is lets say x for now. you have to make that letter by its self.
For Eg.
solve for x
2(x+3)+2(y-2)=3(x-2)
2x+6+2y-4=3x-6
2x+2y+2=3x-6
2x+2y=3x-6-2
2x+2y=3x-8
2x-3x+2y=-8
-x+2y=-8
-x=-2y-8
(-x=-2y-8)/-1
x=2y+8
I hope that this helps you out.
PostPosted: Fri Apr 22, 2005 4:11 pm
by Artist4Jesus89
Thank you it does help
PostPosted: Sun Jun 19, 2005 10:15 am
by lostlamb99
Wait You said Solve for h right?
So if this is the equation: A = h/2 (b + c)
Hmmm! If I recall My Algebra from last Year I think this is how you solve it:
Say you give a value for the other Variables like
A= 18, b= 8, and c= 4
so In order to answer this you have to substitute the value of the other variables.
18= h/2 (8 + 4)
18= h/2 (12)
18= 12h/2
18= 6h
18/6 = 6h/6
h= 3
Let's Check:
18= 2/2 (8+4)
18= 3/2 (12)
18= 18
so I hope this helps. ^_^
