U people need to concentrate a little to understand ...
Xà-5 x^3+2x^2-15x
=lim |x+5|
xà-5 x(x+5)(x-3) X^3+2x^2-15
NOW
x+5 if x>-5
|x+5|= -(x+5) if x<-5
0 if x=-5
Left Hand Limit
Lim _ |x+5|
xà-5 x(x+5)(x-3)
=lim _ -(x+5)
xà-5 x(x+5)(x-3)
=lim _ -1
xà-5 x(x-3)
= -1 = -1
-5(-5-3) -5(-8)
= -1
40
Right Hand Side
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Lim |x+5|
xà-5^+ x(x+5)(x-3)
=lim
xà-5^+ x+5 = 1
x(x+5)(x-3) 40
NOW
Left Hand Side # Right Hand Side
LIMIT DOES NOT EXIST AT X= -5
U people need to concentrate a little to understand ...
please jaldi karian please
kia jaldi karna hai????? ek bonus day bhi hota hai janab.
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assignment on attachment
Question # 2
4x2+4y2+16y-4x=19
4x2+4y2+2(-2)x+2(4y)-19=0
g = -2
f = +8 c = -19
r = f 2 + g2 − c
= (−2)2 + 82 − (−19)
= 4 + 64 + 19
= 87
Radius = 87
Center = (-2, 8)
Question # 3
x ≥ −2
8 − x
⇒ x ≥ −2 ( 8 − x )
⇒ x ≥ −16 + 2 x
⇒ x − 2 x ≥ −16
⇒ − x ≥ −16
⇒ x ≤ 16
⇒ { x x ∈ R x ≤ 16}
plz upload it in word or pdf file it is very dificult to unerstand the words.
plz today is last date of this assignment
oh ho. ok let me do something for you . i guess you still have a bonus day...
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plz upload question NO.1
plz hurry up
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