Mathematics: Find the illegal values of c in the multiplication statement. c²-3c-10/c²+5c-14 · c²-c-2/c²-2c-15 a. c = –7, c = –3, c = –2, and c = 5 b. c = –7, c = –3 c. c = 7, c = 3, c = –2, and c = –5 d. c = –7, c = –3, c = 2, and c = 5

Find the illegal values of c in the multiplication

Ответ:

chiquita77

12.01.2022

The illegal values of c in the multiplication statement
$\frac{c^2-3c-10}{c^2+5c-14}\cdot \frac{c^2-c-2}{c^2-2c-15}$
are the values of c for which
c^2+5c-14=0

and

For
$c^2+5c-14=0 \\ \\ (c-2)(c+7)=0 \\ \\ c-2=0 \ and \ c+7=0 \\ \\ c=2 \ and \ c=-7$

For
$c^2-2c-15=0 \\ \\ (c-5)(c+3)=0 \\ \\ c-5=0 \ and \ c+3=0 \\ \\ c=5 \ and \ c=-3$

Therefore, the illegal values of c in the multiplication statement
$\frac{c^2-3c-10}{c^2+5c-14}\cdot \frac{c^2-c-2}{c^2-2c-15}$
are -7, -3, 2 and 5.

0,0(0 оценок)

Ответ:

20alondra04

08.11.2022

Mathematics

I believe 7 would be B and 8 would be D.

Please let me know if I'm wrong :)

0,0(0 оценок)

Find the illegal values of c in the multiplication statement. c²-3c-10/c²+5c-14 · c²-c-2/c²-2c-15

a. c = –7, c = –3, c = –2, and c = 5
b. c = –7, c = –3
c. c = 7, c = 3, c = –2, and c = –5
d. c = –7, c = –3, c = 2, and c = 5

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Mathematics

Ask an AI advisor a question