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2 years ago

5

Miriam ha dibujado un círculo y luego, con mucha paciencia, ha trazado 2016 diámetros diferentes. ¿En cuántos gajos ha quedado d

ividido?

1 answer:

bearhunter [10]2 years ago

6 0

Answer:

Yeah I got dogs

Step-by-step explanation:

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3 years ago

P and W are twice-differentiable functions with P(7)=2(W(7)). The line y=9+2.5(x-7) is tangent to the graph of P at x=7. The lin

fiasKO [112]

Using the product rule, we have

$m(x) = x W(x) \implies m'(x) = xW'(x) + W(x)$

so that

$m'(7) = 7W'(7) + W(7)$

The equation of the tangent line to <em>W(x)</em> at <em>x</em> = 7 has all the information we need to determine <em>m'</em> (7).

When <em>x</em> = 7, the tangent line intersects with the graph of <em>W(x)</em>, and

<em>y</em> = 4.5 + 2 (7 - 7) ==> <em>y</em> = 4.5

means that this intersection occurs at the point (7, 4.5), and this in turn means <em>W</em> (7) = 4.5.

The slope of this tangent line is 2, so <em>W'</em> (7) = 2.

Then

$m'(7) = 7\cdot2 + 4.5 = \boxed{18.5}$

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2 years ago

In quadrilateral $ABCD$, we have $AB=3,$ $BC=6,$ $CD=4,$ and $DA=4$.

vampirchik [111]

The triangle inequality applies.

In order for ACD to be a triangle, the length of AC must lie between CD-DA=0 and CD+DA=8.

In order for ABD to be a triangle, the length of AC must lie between BC-AB=3 and BC+AB=9.

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3 years ago

A 6th grade teacher can grade 25 homework assignments in 20 minutes. Is this working at a faster rate or slower rate than gradin

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25 ÷ 20 = 1.25
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3 years ago

Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form

Alex787 [66]

Answer:

$y=-\frac{2}{5}x-\frac{1}{5}$

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

m is the slope

b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

$m = -\frac{2}{5}$

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

$y=-\frac{2}{5}x + b$

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,1). When x of the line is -3, y of the line must be 1.
(2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: $y=-\frac{2}{5}x + b$ . $b$ is what we want, the - $-\frac{2}{5}$ is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

(-3,1). y = mx + b or $1=-\frac{2}{5} * -3 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(-3)$ . $b = -\frac{1}{5}$ .
(2,-1). y = mx + b or $-1=-\frac{2}{5} * 2 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points (-3,1) and (2,-1) is $y=-\frac{2}{5}x-\frac{1}{5}$

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2 years ago