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bixtya [17]
3 years ago
15

Please help, thanks if you do :)

Mathematics
1 answer:
Evgen [1.6K]3 years ago
8 0

Answer: No solution

Explanation: It is completely parallel. If it intersects, it has 1 solution. If it is exactly the same, giving the appearance of one line, it has infinite. When it’s parallel, it has none.

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Its infinite solutions because its the same exact thing when you simplify the first half of the problem.

Step-by-step explanation:

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Step-by-step explanation:

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8 0
3 years ago
Evaluate................
wolverine [178]

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

__

2. Same as the first problem.

k(x) = 8x +7q

k(2q² +3q) = 8(2q² +3q) +7q = 16q² +24q +7q = 16q² +31q

_____

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.

7 0
3 years ago
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