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.
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
3 answers
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
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
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.