Please help solve this problem...
Solve the equation for b1
A= 1/2h(b1+b2)
Also do you need to reduce anyhting in the following fraction?
36/8y^4
2007-08-30 09:08:50 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Solving areas of trapezoids, are we?
A = (1/2) * h * (b1 + b2)
2A = h * (b1 + b2)
2A/h = (b1 + b2)
2A/h - b2 = b1
36/8 is an improper fraction. It is 4 4/8 or 4 1/2
(4 1/2) * (y^4) is as simple as it gets
2007-08-30 09:27:11 · answer #1 · answered by PMP 5 · 0⤊ 0⤋
If you mean A = (1/2)*h*(b1 + b2): multiply both sides by 2, then divide both sides by h, then subtract b2 from both sides. Your equation will have been solved for b1.
2007-08-30 09:21:19 · answer #2 · answered by Tony 7 · 0⤊ 0⤋
A = 1/2h(b1 +b2) = 2A = h(b1 + b2) = 2A/h = b1 + b2 = 2A/h - b2 = b1. 36/8y^4 reduces to 9/2y^4. Thanks for the 2 points!
2007-08-30 09:28:23 · answer #3 · answered by Emissary 6 · 0⤊ 0⤋
In the first equation, here are the first two steps to get you going:
1. Multiply both sides by 2
2. Divide both sides by h.
3. ... is a subtraction that I'm sure you can figure out
For the second expression, ask yourself if the numerator and denominator have any common factors. (Hint: 36= 2x2x3x3).
2007-08-30 09:18:58 · answer #4 · answered by richg74 3 · 0⤊ 0⤋
A= 1 / 2h(b1 + b2)
A (b1+b2) = 1/2h
b1+b2 = 1/2Ah
b1 = 1/2Ah - b2
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36 / 8y^4
(9*4) / (2*4) y^4
9 / (2y^4)
-----------------
Solutions:
b1 = 1/(2Ah) - b2
9 / (2 y^4)
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2007-08-30 09:20:50 · answer #5 · answered by Anonymous · 0⤊ 0⤋
36/8y^4
=9/2y^4. ANS.
2007-08-30 09:18:23 · answer #6 · answered by Anonymous · 0⤊ 0⤋
1/2Ah = b1+b2
1/2Ah -b2 = b1
9/2y^4
2007-08-30 09:14:40 · answer #7 · answered by Anonymous · 0⤊ 0⤋