The correct answers are x≠2 and x≠(-2).
Domain restrictions are any points in the domain where the function will not have a value. This basically means that it's a point where x won't work in the function.
For the numerator, any value will work for x. We can have a value of 0 in the numerator, or a positive, negative, or decimal number.
However, the denominator cannot equal 0. This is because a fraction bar represents division, and we cannot divide by 0. The values that make the denominator 0 can be found by:
x²-4=0
Add 4 to both sides:
x²-4+4 = 0+4
x² = 4
Take the square root of both sides:
√x² = √4
x = 2 or x = -2.
Answer:
See below
Step-by-step explanation:
I will assume the given denominator is 3a as there is nothing attached to the statement.
Lets first look at the domain of 2a+b. As it is a polynomial with unknowns a and b, we know it has its domain in all real numbers for both a and b as every polynomial does. For verifying it replace a and b by any real number you can think of.
Domain = a, b belonging to R
Now, if we divide 2a+b by 3a we will have:
(2a+b)/3a
As know we don not have a polynomial we cant state that the domain will still be all the real numbers for sure. We need to go further.
We have a fraction and as every fraction the denominator CAN'T be equal to 0, so 3a MUST be different to 0. This means that a MUST be different to 0. So, now our domain changes:
domain' = a, b belonging to R and a different to 0.
Answer:
D. y = -5^x – 27
Step-by-step explanation:
Find the negative reciprocal of the slope of the original line and use the point-slope formula y
−
y
1
=
m
(
x
−
x
1
) to find the line perpendicular to −
x
+
5
y
=
14
.
y
=
−
5
x
−
27 PERPENDICULAAR
...................................................................................................................................................
Find the slope of the original line and use the point-slope formula y
−
y
1
=
m
(
x
−
x
1
) to find the line parallel to −
x
+
5
y
=
14
.
y
=
1
/5
x
−
1 PARALLEL
What are you asking anyways hope you get a good grade
