Respuesta :
The options have not been given but three correct alternatives have been presented as the solution.
We have given,
[tex]\frac{c^2-4}{c+3} /\frac{c+2}{3(c^2-9)}[/tex]
Applying the rule a^2-b^2 to the numerator of the first term and the denominator of the second term,
What is the rule of a^2-b^2?
[tex](a^2-b^2)=(a-b)(a+b)[/tex]
By applying the above formula we have:
[tex]=\frac{c^2-2^2}{c+3} /\frac{c+2}{3((c+3)(c-3)}[/tex]
[tex]=\frac{(c-2)(c+2)}{c+3} /\frac{c+2}{3((c+3)(c-3)}[/tex]
Therefore the solution is,
[tex]=(c-2) \times 3(c-3)\\\\=3(c^2-5c+6)\\\\=3c^2-15c+18[/tex]
To learn more about the expression visit:
https://brainly.com/question/723406
Answer:
C.) StartFraction c squared minus 4 Over c + 3 EndFraction times StartFraction 3 (c squared minus 9) Over c + 2 EndFraction"/>
Step-by-step explanation:
Which expression is equivalent to StartFraction c squared minus 4 Over c + 3 EndFraction divided by StartFraction c + 2 Over 3 (c squared minus 9) EndFraction?
StartFraction c + 3 Over c squared minus 4 EndFraction times StartFraction c + 2 Over 3 (c squared minus 9) EndFraction
StartFraction c squared minus 4 Over c + 3 EndFraction divided by StartFraction 3 (c squared minus 9) Over c + 2 EndFraction
StartFraction c squared minus 4 Over c + 3 EndFraction times StartFraction 3 (c squared minus 9) Over c + 2 EndFraction
StartFraction c squared minus 4 Over c + 3 EndFraction times StartFraction c + 2 Over 3 (c squared minus 9) EndFraction
