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krek1111 [17]
3 years ago
15

Which functions have removable discontinuities (holes)? Check all of the boxes that apply.

Mathematics
2 answers:
Naya [18.7K]3 years ago
4 0

Answer:

a b d

Step-by-step explanation:

Zolol [24]3 years ago
3 0

Answer:

Option A,B,D

Step-by-step explanation:

For f(x) = (x-1)/(x²-1)

f(x) = (x-1)/(x-1)(x+1)

Therefore hole at x = 1

For f(x) = (x²-9)/(x²+7x+12) = (x-3)(x+3)/(x²+4x+3x+12) = (x-3)(x+3)/(x+3)(x+4)

Common factor (x+3) in numerator and denominator so

Function has hole at x = -3

For f(x) = (x² + 4x + 4)/(x² + 2x - 8) = (x+2)²/(x² +4x -2x -8) = (x+2)²/(x+4)(x-2)

This function has no discontinuity.

For f(x) = (x +7)/(x²+5x-14) = (x+7)/(x² + 7x -2x -14) = (x+7)/(x-2)(x+7)

This function has a hole at x = -7

Option A,B,D have the removable discontinuity.

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Answer:

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Step-by-step explanation:

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Check the picture below.

to graph an inequality, you pretty much first off, need to graph the "equality" or equation, so for y < 3x − 6, first off, graph the line <span>y = 3x − 6.

and then you do a "true" or "false" check.

now, if you look at the line below, it splits the grid in two sections, so, let's check a point in either section, if one is false, the other is true and the other way around.

so, let's check the region on the left-hand-side, hmmmm say point 0,0, the origin.

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because the point 0,0 yielded a false inequality, that region is the "false region" and thus not shaded, therefore, the other side must be the "true region", and thus we shade that then.

we can run a quick check on that btw, say on point 4,4

y =4,   x = 4  thus

4 < 3(4) - 6

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