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Sav [38]
3 years ago
12

I don’t know how to do this ? Calculate the slope of the line.

Mathematics
1 answer:
Marrrta [24]3 years ago
4 0

Answer:

The slope of the line is -2/1=  -2

Step-by-step explanation:

To get the answer you would have to do rise over run the rise is two and the run which is the bottom of the line would be one and sinse the line is going down it would be negative

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3 Ximena draws an X shape in the coordinate plane as shown. Draw the images of
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When a shape is rotated, it must be rotated around a point.

<em>See attachment for the image of each rotation.</em>

To do this, the top coordinates of the X shape will be transformed using the appropriate rotation rule; the same rule will then be applied to the other parts of the X shape.

The top coordinates of the X shape are:

A = (0,0)

B = (1,0)

C = (3,0)

D =(4,0)

For 90 degrees counterclockwise rotation, the rule is:

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So, we have:

A' = (0,0)

B' = (0,1)

C' = (0,3)

D' = (0,4)

For 180 degrees rotation, the rule is:

(x,y) \to (-x,-y)

So, we have:

A' = (0,0)

B' = (-1,0)

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D' = (-4,0)

For 270 degrees counter rotation, the rule is:

(x,y) \to (y,-x)

So, we have:

A' = (0,0)

B' = (0,-1)

C' = (0,-3)

D' = (0,-4)

See attachment for the image of each rotation

Read more about rotations at:

brainly.com/question/1571997

3 0
3 years ago
Help me and fast! And please show work!
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What is the horizontal asymptote for y(t) for the differential equation dy dt equals the product of 2 times y and the quantity 1
marta [7]
First, we need to solve the differential equation.
\frac{d}{dt}\left(y\right)=2y\left(1-\frac{y}{8}\right)
This a separable ODE. We can rewrite it like this:
-\frac{4}{y^2-8y}{dy}=dt
Now we integrate both sides.
\int \:-\frac{4}{y^2-8y}dy=\int \:dt
We get:
\frac{1}{2}\ln \left|\frac{y-4}{4}+1\right|-\frac{1}{2}\ln \left|\frac{y-4}{4}-1\right|=t+c_1
When we solve for y we get our solution:
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To find out if we have any horizontal asymptotes we must find the limits as x goes to infinity and minus infinity. 
It is easy to see that when x goes to minus infinity our function goes to zero.
When x goes to plus infinity we have the following:
$$\lim_{x\to\infty} f(x)$$=y=\frac{8e^{c_1+\infty}}{e^{c_1+\infty}-1} = 8
When you are calculating limits like this you always look at the fastest growing function in denominator and numerator and then act like they are constants. 
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3 0
3 years ago
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