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grrrl9
11-01-2013, 08:03 PM
Hi, I'm a bit confused with how to go about solving part of a word problem......

The formula: C = 4*d-1/3 b

when d = 23245 and b = 13.5

C = 4(23,245) -1/3 (13.5)

C = 4(.035)(13.5)

C = .14(13.5)

I already solved the first part for C and came up with 1.89 (my work is shown above).

the second part asks us to solve for d, but the example given is a bit confusing and gives the following formula: D = 1.1g-1/4 H

I am not sure which number I am supposed to use (in place of the 1.1 given in the example) to solve for d. I think that it is one of the numbers from the steps taken to solve for C in the first part of this problem, I am just a bit confused on how to determine which value to use. Please help point me in the right direction!

JeffM
11-01-2013, 09:00 PM
Hi, I'm a bit confused with how to go about solving part of a word problem......

The formula: C = 4*d-1/3 b

when d = 23245 and b = 13.5

C = 4(23,245) -1/3 (13.5)

C = 4(.035)(13.5)

C = .14(13.5)

I already solved the first part for C and came up with 1.89 (my work is shown above).

the second part asks us to solve for d, but the example given is a bit confusing and gives the following formula: D = 1.1g-1/4 H

I am not sure which number I am supposed to use (in place of the 1.1 given in the example) to solve for d. I think that it is one of the numbers from the steps taken to solve for C in the first part of this problem, I am just a bit confused on how to determine which value to use. Please help point me in the right direction!
It would help a lot if you gave the problem exactly instead of paraphrasing. It would also help if you explained what g and H are.

C = 4*d^{-(1/3)}b = \dfrac{4b}{\sqrt[3]{d}} \implies \sqrt[3]{d} = \dfrac{4b}{C} \implies d = \dfrac{64b^3}{C^3}.