plz help with my precal question!!!?

Question

ok heres the question, i need to know how to work it out too....

You obtain the following answer to a homework problem, 1/10(2x-1)to the 5/2 + 1/6(2x-1)to the 3/2......
the back of the book has an answer of 1/15(2x-1)to the 3/2 then that times (3x-1). Are these two answers equivalent? if so, show how the second answer can be obtained from the first.

*note: when it shows ex. 1/2 its a fraction not division!

thx

they dont have to equal..... but if they are, i need to know how.....

Answer 1

Well, (2x-1)^(5/2) is, of course, the same thing as (2x-1)*(2x-1)^(3/2).

Substituting that into your answer, you get:
(1/10)*(2x-1)*(2x-1)^(3/2) + (1/6)*(2x-1)^(3/2)
((x/5) - (1/10) + (1/6)) * (2x-1)^(3/2)
((x/5) - (3/30) + (5/30)) * (2x-1)^(3/2)
((x/5) + (2/30)) * (2x-1)^(3/2)
((x/5) + (1/15)) * (2x-1)^(3/2)

Nope, looks to me like you made a mistake somewhere - your answer is not equivalent to the book answer.

Answer 2

The answer in the back of the book is correct.

(1/10)(2x-1)^(5/2) + (1/6)(2x-1)^(3/2)
=(3/30)(2x-1)^(5/2) + (5/30)(2x-1)^(3/2)
=(3(2x-1)^(5/2) +5(2x-1)^(3/2)/30
=(3(2x-1)(2x-1)^(3/2) +5(2x-1)^(3/2)/30
=((3(2x-1)+5)(2x-1)^(3/2))/30
=((6x+2)(2x-1)^(3/2))/30
=((3x+1)(2x-1)^(3/2))/15
=(3x+1)(2x-1)^(3/2))/15

or = (1/15)(3x+1)(2x-1)^(3/2)
Which is, I believe, what you said the book answer is.

Answer 3

First it is incredibly not precal, it incredibly is trigonometry 2nd its stressful to confirm what you're asserting. one million. 2. 3. Use pythagorean theorem and merely dvide the factors. upload multiply or exchange the equation whether you like afterwards. or merely play with the factors. 4. exchange the values slightly. 4. (sinX/cosX)/(hyp/opp)-yadi yada. ((opp/hyp)/(upload/hyp))*opp/hyp = cosX a similar happens to the different ingredient. you fee is 0 cosX-cosX=0 genuine looking? 5. can not examine it.

Answer 4

um............. its 13 /10 trust me on this one