The expression will be d. 9-7i
$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
Answer:
m ≠1 ( all m in R except 1 )
Step-by-step explanation:
hello :
mx − y + 3 = 0.....(*)
(2m − 1)x − y + 4 = 0 ....(**)
multiply (*) by : -1 you have : -mx+y-3=0 ....(***)
(2m − 1)x − y + 4 = 0 ....(**)
add(***) and(**) : -mx+ (2m − 1)x+1 =0
(2m-m-1)x+1=0
(m-1)x = -1
this system have no solution if : m-1≠0 means : m ≠1
9514 1404 393
Answer:
C. x^2 + 3
Step-by-step explanation:
Substitute for f(x) and g(x) and simplify.
(f -g)(x) = f(x) -g(x)
(f -g)(x) = (2x^2 +2) -(x^2 -1) = 2x^2 +2 -x^2 +1
(f -g)(x) = x^2 +3
Answer:
Step-by-step explanation:
Look at the picture.
ΔADC and ΔCDB are similar. Therefore the sides are in proportion:
We have
Substitute:
<em>cross multiply</em>
For x use the Pythagorean theorem:
