<h3>
Answer: Choice A</h3>
- Domain: x > 4
- Range: y > 0
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Explanation:
We want to avoid having a negative number under the square root. Solving
leads to 
So it appears the domain could involve x = 4 itself; however, if we tried that x value, then we'd get a division by zero error.
So in reality, the domain is x > 4.
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The range of y = sqrt(x) is the set of positive real numbers. So y > 0 is the range for this equation. Shifting left and right does not affect the range, so the range of y = sqrt(x-4) is also y > 0.
We are dividing a positive number (3) over some positive number in the denominator. Overall, the expression
is positive because positive/positive = positive.
Therefore, the range of the given equation is y > 0
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The graph is shown below. We have a vertical asymptote at x = 4 and a horizontal asymptote at y = 0. The green curve is fenced in the upper right corner (northeast corner).
Answer:
943,281
Step-by-step explanation:
Just move the 3 up a value
If it's in the ten's place, then move it to the hundred's place value.
If it's in the hundred's place, then move it to thousand's place value.
So on and so on...
(f+g)(x)=5x-6+x^2-4x-8
A.(f+g)(x)=x^2+x-14 is correct:)
2x + 5 = 11
2x = 6
x = 3
y + 4 = 2x + 4
y + 4 = 2(3) + 4
y + 4 = 10
y = 6
answer
x = 3 and y = 6
-x^3 + x^2/3 + 5 - 13/x
I don't know if this is correct