The radius of a circle can be found by dividing the circumference of the circle by 6.28. What is the radius of a circle that has
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Answer:
Step-by-step explanation:
Given:
Gisselle has 1/2 of a candy bar.
She role off 1/2 of what she had and split it into three equal pieces.
Question asked:
What fraction of a candy bar was each piece?
Solution:
Gisselle has =
She role off =
As she split it into three equal pieces, we will divided it by 3, we get,
Therefore, fraction of a candy bar of each piece is
I'm assuming the function is f(x) = (2x+8)/(x^2+5x+6). If so, make sure to use parenthesis to indicate that you're dividing all of "2x+8" over all of "x^2+5x+6" as one big fraction. Otherwise, things are ambiguous and it leads to confusion.
Side Note: x^2 means "x squared"
Factor the numerator: 2x+8 = 2(x+4)
Factor the denominator: x^2+5x+6 = (x+2)(x+3)
There are no common factors between the numerator and denominator. So there is nothing to cancel out.
Recall that you cannot divide by zero. Something like 1/0 is undefined.
We need to find the x values that cause the denominator to be zero.
Set the denominator equal to zero and solve for x
x^2+5x+6 = 0
(x+2)(x+3) = 0
x+2 = 0 or x+3 = 0
x = -2 or x = -3
The x values x = -2 or x = -3 will lead to the denominator being zero. This means that the vertical asymptotes are x = -2 or x = -3 as shown by the blue dashed vertical lines in the attached image.
Side Note: x^2 means "x squared"
Factor the numerator: 2x+8 = 2(x+4)
Factor the denominator: x^2+5x+6 = (x+2)(x+3)
There are no common factors between the numerator and denominator. So there is nothing to cancel out.
Recall that you cannot divide by zero. Something like 1/0 is undefined.
We need to find the x values that cause the denominator to be zero.
Set the denominator equal to zero and solve for x
x^2+5x+6 = 0
(x+2)(x+3) = 0
x+2 = 0 or x+3 = 0
x = -2 or x = -3
The x values x = -2 or x = -3 will lead to the denominator being zero. This means that the vertical asymptotes are x = -2 or x = -3 as shown by the blue dashed vertical lines in the attached image.