-153,551, -245, -15.6, 15,410
Least to Greatest --->
Let x represent the domain
Let y represent the range.
Assume that y =a*b⁻ˣ
Set x=0:
a*b⁰ = 32
a = 32
Set x=1:
32*b⁻¹ = 24
32 = 24b
b = 32/24
Set x=2:
d = 32*(32/24)⁻²
= 32*(24/32)²
= (32*24*24)/(32*32) = 24²/32 = 576/32
= 18
Answer: 18
Hello,
Use the factoration
a^2 - b^2 = (a - b)(a + b)
Then,
x^2 - 81 = x^2 - 9^2
x^2 - 9^2 = ( x - 9).(x + 9)
Then,
Lim (x^2- 81) /(x+9)
= Lim (x -9)(x+9)/(x+9)
Simplity x + 9
Lim (x -9)
Now replace x = -9
Lim ( -9 -9)
Lim -18 = -18
_______________
The second method without using factorization would be to calculate the limit by the hospital rule.
Lim f(x)/g(x) = lim f(x)'/g(x)'
Where,
f(x)' and g(x)' are the derivates.
Let f(x) = x^2 -81
f(x)' = 2x + 0
f(x)' = 2x
Let g(x) = x +9
g(x)' = 1 + 0
g(x)' = 1
Then the Lim stay:
Lim (x^2 -81)/(x+9) = Lim 2x /1
Now replace x = -9
Lim 2×-9 = Lim -18
= -18
What’s the original question and I can help!
Answer:
x = -3 and x = -3/2
Step-by-step explanation:
After writing down the polynomial, split it; put a line between 3x^2 and -18x. Look and 2x^3 + 3x^2 and -18x - 27 separately and factor them both:
p(x) = 2x^3 + 3x^2 <u>- 18x -27</u>
p(x) = x^2(2x+3) <u>-9(2x+3)</u>
Now notice how x^2 and -9 have the same factor (2x+3). That means x^2 and -9 can go together:
p(x) = (x^2 - 9)(2x+3)
Factor it once more because there's a difference of squares:
p(x) = (x+3)(x-3)(2x+3)
Now just plug in whatever makes the each bracket equal 0:
x = -3, x = 3, and x = -3/2
Those are your zeros.
