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n200080 [17]
3 years ago
7

Can someone plz help me???

Mathematics
2 answers:
Ksenya-84 [330]3 years ago
6 0

Answer:

y = -x - 1

Step-by-step explanation:

y = mx + b

m = slope

b = y-intercept

On the graph, the point on the y-axis is the y-intercept. That point is (0, -1). So the y-intercept is -1.

To find the slope, find two points and put it in the slope equation.

Point A: (0, -1)

Point B: (-1, 0)

m = \frac{y_{2}-y_{1}}{x_{2}-x{1}}

Point A: (x_{1}, y_{1})

Point B: (x_{2}, y_{2})

m = \frac{0 - (-1)}{-1 - 0}

m = \frac{1}{-1}

m = -1

y = -1x - 1

In other words, y = -x - 1

Hope this helped! <3

Kitty [74]3 years ago
4 0

Answer:  y = -1x - 1

This means -1 goes in the first box, and -1 goes in the second box.

y = -1x - 1 is the same as y = -x - 1

========================================================

Explanation:

The two points (0,-1) and (1,-2) are on the red diagonal line.

Let's use the slope formula to compute the slope

m = (y2-y1)/(x2-x1)

m = (-2-(-1))/(1-0)

m = (-2+1)/(1-0)

m = -1/1

m = -1

The slope is -1.

The y intercept is -1 because the red line crosses the y axis at -1.

Since the y intercept is -1, this means b = -1.

With m = -1 and b = -1, we go from y = mx+b to y = -1x + (-1). This is the same as y = -1x - 1

Furthermore, that equation simplifies to y = -x - 1.

Side note: the slope and y intercept aren't always the same value.

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Answer:

x = 1

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Step-by-step explanation:

A*B= (4y-8 3x-3)

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5 0
3 years ago
What is the range for the following function? y=1/(x+2)+3
LUCKY_DIMON [66]

Answer:

\large \boxed{\text{C) }\{y: y \in \mathbb{R}, y \ne 3\} }

Step-by-step explanation:

The range is the spread of the y-values (minimum to maximum distance travelled).

The graph of your function is a hyperbola shifted two units left and three units up from the origin.

There is a vertical asymptote at x = -2, so y does not exist when x = -2.  However,

\displaystyle \lim_{x \rightarrow -{2}^{+}}f(x) = \lim_{x \rightarrow -{2}^{+}}\left (\dfrac{1}{x+2}+3 \right ) = 0 + 3 = 3\\\\\lim_{x \rightarrow -{2}^{-}}f(x) = \lim_{x \rightarrow -{2}^{-}}\left (\dfrac{1}{x+2}+3 \right ) = 0 + 3 = 3

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\displaystyle \lim_{x \rightarrow -{2}}f(x) = 3

The graph below shows the asymptotes of your function.

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In set builder notation, the range is  

\large \boxed{\mathbf{\{y: y \in \mathbb{R}, y \ne 3\} }}

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