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Mekhanik [1.2K]
3 years ago
5

If you could help me out, that would be great. I will give brainliest

Mathematics
1 answer:
laiz [17]3 years ago
8 0

Answer:

y-3=-2(x+2)

Step-by-step explanation:

Point-slope form is y-y_1=m(x-x_1)

m= slope, y_1=3,  x_1=-2

slope= \frac{rise}{run}=\frac{-2}{1} =-2

y-3=-2(x-(-2))

y-3=-2(x+2)

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Graphing help please - fairly simple, image attached.
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To figure out if a graph is a function, you can use this thing called the vertical line test. in case you're unfamiliar, it's basically where you just imagine a vertical line going from left to right on the graph. if it crosses the function in two places, it's incorrect. i suggest looking up a video that shows you a visual representation of the vertical line test if you've never heard of it; it's fairly simple.

A is a function because the graph passes the vertical line test. if you imagine a vertical line and drag it from left to right across the graph, the linear function graphed in choice A does not have two x values at the same spot.

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C is a function. it passes the vertical line test, even though it looks a little strange. at no point does it intersect the vertical line twice.

D is a function. again, it doesn't intersect the vertical line twice.

now, to determine if a function is a one-to-one function, it must also pass the horizontal line test. this means that it doesn't intersect at two points horizontally as well. test that out on choices A, C, and D.

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3 years ago
EVERYONE BE HONEST!!<br> Doesn't Ariana look like 2 different people?? (Hair up and hair down)
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Answer:

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Step-by-step explanation:

a) It can be convenient to use the square root function as one being defined only for values greater than or equal to some number. Here, we need to shift that function 3 units to the left, so its domain is [-3, ∞).

  f(x)=\sqrt{x+3}

__

b) One way to put holes in a function is to put vertical asymptotes there. This function can be defined to have vertical asymptotes at x=2 and x=4 so those values are excluded from the domain.

  f(x)=\dfrac{1}{(x-2)(x-4)}

3 0
3 years ago
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