Asked by Sabrina
I posted a few questions and then went back and posted how I thought they were supposed to be answered. Could someone please locate my earlier questions and tell me if I am on the right track?
Answers
Answered by
Reiny
I answered some of them
click on your name above and you will see your posts
click on your name above and you will see your posts
Answered by
Sabrina
Thank you Reiny so much for you help. Can you tell me if I am on the right track with this problem: 4c^2/4c^2-8c+4 x 4c-4/2c When solving this am I supposed to find the gcf or lcm to solve> would the lcm be 2 and the gcf be 16? I don't know if I am even going down the right road here?
Answered by
Reiny
Use brackets to show when the denominator ends
I think you meant
4c^2/(4c^2-8c+4)( 4c-4)/2c , which is
= 4c^2/(4(c^2 - 2c + 1)(4)(c-1)/(2c)
= 4c^2/(4(c- 1)^2(4)(c-1)/(2c)
now things cancel rather nicely
final result
2c/(c-1) , c ≠ 1,0
I think you meant
4c^2/(4c^2-8c+4)( 4c-4)/2c , which is
= 4c^2/(4(c^2 - 2c + 1)(4)(c-1)/(2c)
= 4c^2/(4(c- 1)^2(4)(c-1)/(2c)
now things cancel rather nicely
final result
2c/(c-1) , c ≠ 1,0
Answered by
Sabrina
THANKS!!!!!! Just one question, when I follow these steps and get to the result is it both 2c/(c-1), c‚ 10 or am I reading my problem wrong and the answer is just c‚ 10
Answered by
Reiny
The final answer is
2c/(c-1)
the part after it is the restriction.
you are reading it wrong, it does not have 10 in it.
it says, c cannot be equal to 1 or c cannot be equal to 0
Since we cannot divide by zero, those values would make the denominators we canceled equal to zero.
2c/(c-1)
the part after it is the restriction.
you are reading it wrong, it does not have 10 in it.
it says, c cannot be equal to 1 or c cannot be equal to 0
Since we cannot divide by zero, those values would make the denominators we canceled equal to zero.