Answer:
As x decreases, h increases (but never touches)
As x increases, h decreases (but never touches)
Step-by-step explanation:
If you graphed the exponential, you would be able to see that it is stuck in the 3rd quadrant.
Domain: (negative infinity, 1)
Range: (negative infinity, 0)
Answer:
297.60 & 1497.60
Step-by-step explanation:
I had the question before and got it right
Answer:
h(8q²-2q) = 56q² -10q
k(2q²+3q) = 16q² +31q
Step-by-step explanation:
1. Replace x in the function definition with the function's argument, then simplify.
h(x) = 7x +4q
h(8q² -2q) = 7(8q² -2q) +4q = 56q² -14q +4q = 56q² -10q
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2. Same as the first problem.
k(x) = 8x +7q
k(2q² +3q) = 8(2q² +3q) +7q = 16q² +24q +7q = 16q² +31q
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Comment on the problem
In each case, the function definition says the function is not a function of q; it is only a function of x. It is h(x), not h(x, q). Thus the "q" in the function definition should be considered to be a literal not to be affected by any value x may have. It could be considered another way to write z, for example. In that case, the function would evaluate to ...
h(8q² -2q) = 56q² -14q +4z
and replacing q with some value (say, 2) would give 196+4z, a value that still has z as a separate entity.
In short, I believe the offered answers are misleading with respect to how you would treat function definitions in the real world.
Answer:
Step-by-step explanation:
<h2>Equation of the line in slope intercept form:</h2>
Here m is the slope or gradient and b is the y-intercept.
Write the given equation in slope-intercept form.
2x + 3y = 6
3y = -2x + 6
