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damaskus [11]
3 years ago
6

Find the slope of the line that passes through the given points. (10,-15) (13,-17)

Mathematics
1 answer:
Studentka2010 [4]3 years ago
4 0
M=-2/3

This is because it’s the slope I got and I don’t know how to explane it
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A square is a rhombus <br> A. Always <br> B. Sometimes<br> C. Never
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Always. A rhombus is a quadrilateral with equal sides. A square will always have equal sides.
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Wittaler [7]

Answer:

C. x = -1 ; x ~8.1

Step-by-step explanation:

First we graph f(x)= -\sqrt{x+2} -3

Here we have x+2 under the square root. x+2>=0, x>=-2

So we take some x  values greater than -2  and find out f(x)  to make a table for graphing

x        f(x)

-2       -\sqrt{-2+2} -3=-3

-1        -4

2        -5

Graph it  and extent the graph

Now we graph g(x)= -2|x-3| + 4

Given equation is in the form of g(x) = a|x-h| + k

where (h,k) is the vertex

here h= 3  and k = 4, so vertex is (3,4)

Now make a table, pick some numbers for x  less than and greater than vertex

x     y

-1     -2|-1-3| + 4=-4

0    -2|0-3| + 4=-2

3    4

6     -2

Both graphs are attached below

The graph of f(x)  and g(x) interests  at x= -1  and x=8.1

3 0
3 years ago
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The function y =3 square root –x- 3 is graphed only over the domain of {x | –8 &lt; x &lt; 8}. What is the range of the graph?
Flauer [41]
Using a graphing tool, we can see that the whole range of the function

is  <span>y</span><span> ∈ [0,∞)

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 </span>
{x | –8 < x < 8}.

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5 0
3 years ago
Write an equation for an ellipse centered at the origin, which has foci at (\pm5,0)(±5,0)(, plus minus, 5, comma, 0, )and vertic
Virty [35]

Answer:

The equation of ellipse is

\frac{x^2}{41}+\frac{y^2}{16}=1

Step-by-step explanation:

The equation of an ellipse is

\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1

where(h,k) is the center and c is distance from the center to the foci is given by a^2-b^2=c^2. a is the distance from the center to the vertices and b is the distance from the center to the co-vertices.

The center of the ellipse is the mid-point of the vertices.

The mid point of the vertices (\pm\sqrt{41},0) is

=(\frac{\sqrt{41}+(-\sqr{41})}2,\frac{0+0}2)

=(0,0)

a is the distance between the center and the vertices.

So, a=\sqrt{(0-\sqrt{41})^2+(0-0)^2}

        =\sqrt{41}

c is the distance between the center and the foci.

So, c=\sqrt{(0-5)^2+(0-0)^2

         =5

a^2-b^2=c^2

\Rightarrow (\sqrt41)^2-b^2=5^2

\Rightarrow b^2=(\sqrt41)^2- 5^2

\Rightarrow b^2=41-25

\Rightarrow b^2=16

The equation of ellipse is

\frac{(x-0)^2}{(\sqrt{41})^2}+\frac{(y-0)^2}{16}=1

\Rightarrow \frac{x^2}{41}+\frac{y^2}{16}=1

7 0
3 years ago
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