We have that
<span>(c-4)/(c-2)=(c-2)/(c+2) - 1/(2-c)
</span>- 1/(2-c)=-1/-(c-2)=1/(c-2)
(c-4)/(c-2)=(c-2)/(c+2)+ 1/(c-2)------- > (c-4)/(c-2)-1/(c-2)=(c-2)/(c+2)
(c-4-1)/(c-2)=(c-2)/(c+2)---------------- > (c-5)/(c-2)=(c-2)/(c+2)
(c-5)/(c-2)=(c-2)/(c+2)------------- > remember (before simplifying) for the solution that c can not be 2 or -2
(c-5)*(c+2)=(c-2)*(c-2)------------------ > c²+2c-5c-10=c²-4c+4
-3c-10=-4c+4----------------------------- > -3c+4c=4+10----------- > c=14
the solution is c=14
the domain of the function is (-∞,-2) U (-2,2) U (2,∞) or
<span>all real numbers except c=-2 and c=2</span>
(-2,8) (1,2)
m=2-8/1-(-2)
m=-6/3
m= -2
y=-2x+b
(-2,8)
8=-2(-2)+b
8=-4+b
12=b
b=12
y=-2x+12
Hope this helps!
Answer:
18 remainder 15
Step-by-step explanation:
Interquartile Range is the answer
