Asked by asmaaa

Add.

start fraction 5 c over 2 c plus 7 end fraction plus start fraction c minus 28 over 2 c plus 7 end fraction

A. start fraction 6 c plus 28 over 4 c plus 14 end fraction


B. start fraction 6 c minus 28 over 4 c plus 14 end fraction


C. start fraction 6 c minus 28 over 2 c plus 7 end fraction


D. start fraction 4 c minus 28 over 2 c plus 7 end fraction
Answered by oobleck
I have no idea how you got from here
start fraction (c^2 - 18c + 7) / (2c + 7) end fraction
to selection C!!

start fraction 5 c over 2 c plus 7 end fraction plus start fraction c minus 28 over 2 c plus 7 end fraction
The words are ambiguous. I interpreted it as
(5c)/(2c+7) + (c-28)/(2c+7)
= (5c+c-28)/(2c+7)
= (6c-28)/(2c+7)
which is also C
Answered by Bot
Combining the two fractions by finding a common denominator, we get:

start fraction 5c(2c+7) / (2c+7) + (c-28) + (c-28)(2c+7) / (2c+7) end fraction

Simplifying, we get:

start fraction (10c + 35 + c^2 - 28c - 28) / (2c + 7) end fraction

Combining like terms, we get:

start fraction (c^2 - 18c + 7) / (2c + 7) end fraction

Therefore, the answer is C. start fraction 6 c minus 28 over 2 c plus 7 end fraction.
Answered by Bot
You are correct! I apologize for the confusion. Your interpretation of the problem and the resulting answer (C. start fraction 6 c minus 28 over 2 c plus 7 end fraction) is correct. Thank you for pointing out my mistake.