Answer:
1.125
Step-by-step explanation:
Answer:
The correct option is A.
Step-by-step explanation:
Domain:
The expression in the denominator is x^2-2x-3
x² - 2x-3 ≠0
-3 = +1 -4
(x²-2x+1)-4 ≠0
(x²-2x+1)=(x-1)²
(x-1)² - (2)² ≠0
∴a²-b² =(a-b)(a+b)
(x-1-2)(x-1+2) ≠0
(x-3)(x+1) ≠0
x≠3 for all x≠ -1
So there is a hole at x=3 and an asymptote at x= -1, so Option B is wrong
Asymptote:
x-3/x^2-2x-3
We know that denominator is equal to (x-3)(x+1)
x-3/(x-3)(x+1)
x-3 will be cancelled out by x-3
1/x+1
We have asymptote at x=-1 and hole at x=3, therefore the correct option is A....
Answer:c>2
Step-by-step explanation:
-6c (divide) (-6) > -12 (divide) (-6)
c>-12 (divide) (-6)
c>12 (divide) 6
c> 2
Answer:
I Do
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>Sides of the picture added margin:</u>
<u>Solution</u>:
- A(margin) = Total area - A(picture)
- (8 + 2x)(5 + 2x) - 5*8 = 30
- 4x² + 10x + 16x + 40 - 40 - 30 = 0
- 4x² + 26x - 30 = 0
- 2x² + 13x - 15 = 0
- x = (-13 + √(13² + 2*4*15))/4
- x = (-13 + 17)/4
- x = 1 cm
Note. The other root is ignored as negative.
