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Gwar [14]
3 years ago
10

Ok guys now that you have had a good nights sleep, HElP!ME!FIGURE!THIS!OUT!10 points!

Mathematics
2 answers:
amid [387]3 years ago
8 0

I could never get attached

When I start to feel unattached

Somehow always end up feelin' bad

Baby, I am not your dad (no)

It's not all you want from me

I just want your company

Girl, it's obvious, elephant in the room

And we're a part of it, don't act so confused

And you love startin' it, now I'm in a mood

Now we arguin' in my bedroom

eimsori [14]3 years ago
7 0

Answer:

3:5

4:4

15:7

Step-by-step explanation:

they only gave you 1 equation that is equal to 8 so all the others must be equivalent to 8

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Write the equation of a line that passes through two points (-1, 0) and (4, -5).
sleet_krkn [62]

Answer:

<u>y = -1 (x) -1</u>

Step-by-step explanation:

If you plug in the numbers from the coordinates into the equation, you will find that it's true.

-1 is x and 0 is y. Plug in the #s like so:

y = -1 (x) -1

0 = -1 (-1) - 1

0 = 1 - 1

0 = 0

Therefore, the equation above represents the line that passes through both these points.

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3 years ago
If the relative frequencies are 0.48 and 0.52, which
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Answer:The Last One

Step-by-step explanation:

3 0
2 years ago
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Show that every triangle formed by the coordinate axes and a tangent line to y = 1/x ( for x &gt; 0)
vfiekz [6]

Answer:

Step-by-step explanation:

given a point (x_0,y_0) the equation of a line with slope m that passes through the  given point is

y-y_0 = m(x-x_0) or equivalently

y = mx+(y_0-mx_0).

Recall that a line of the form y=mx+b, the y intercept is b and the x intercept is \frac{-b}{m}.

So, in our case, the y intercept is (y_0-mx_0) and the x  intercept is \frac{mx_0-y_0}{m}.

In our case, we know that the line is tangent to the graph of 1/x. So consider a point over the graph (x_0,\frac{1}{x_0}). Which means that y_0=\frac{1}{x_0}

The slope of the tangent line is given by the derivative of the function evaluated at x_0. Using the properties of derivatives, we get

y' = \frac{-1}{x^2}. So evaluated at x_0 we get m = \frac{-1}{x_0^2}

Replacing the values in our previous findings we get that the y intercept is

(y_0-mx_0) = (\frac{1}{x_0}-(\frac{-1}{x_0^2}x_0)) = \frac{2}{x_0}

The x intercept is

\frac{mx_0-y_0}{m} = \frac{\frac{-1}{x_0^2}x_0-\frac{1}{x_0}}{\frac{-1}{x_0^2}} = 2x_0

The triangle in consideration has height \frac{2}{x_0} and base 2x_0. So the area is

\frac{1}{2}\frac{2}{x_0}\cdot 2x_0=2

So regardless of the point we take on the graph, the area of the triangle is always 2.

6 0
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Idk I’m sorry I just need points.
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