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2 years ago

I need help on this, please. thanks

Mathematics

2 answers:

konstantin123 [22]2 years ago

6 0

In the picture are the steps I used to solve it.

koban [17]2 years ago

4 0

Y=4/5x-18/5

The points are (-3,-6)(2,-2)
First find the slope
y2-y1/x2-x1
-2-(-6)/2-(-3)
-2+6/2+3
4/5 slope (m)

y-y1=m(x-x1)
y-(-6)=4/5(x-(-3))
y+6=4/5(x+3)
y+6=4/5x+12/5
y=4/5x+12/5-6
y=4/5x-18/5

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Find Aut( 15). Use the Fundamental Theorem of Abelian Groups to express this group as an external direct product of cyclic group

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Answer:

The solution is attached in the picture below

Step-by-step explanation:

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2 years ago

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Right the slope intercept of the equation of the line through the given point with the given slope through (2, 1) slope = 5/2 m=

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3 years ago

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2 years ago

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3 years ago

please show on graph (with x and y coordinates) state where the function x^4-36x^2 is non-negative, increasing, concave up

babunello [35]

Answer:

$y'' =12x^2 -72=0$

And solving we got:

$x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}$

We can find the sings of the second derivate on the following intervals:

$(-\infty$ Concave up

$x=-\sqrt{6}, y =-180$ inflection point

$(-\sqrt{6}$ Concave down

$x=\sqrt{6}, y=-180$ inflection point

$(\sqrt{6}$ Concave up

Step-by-step explanation:

For this case we have the following function:

$y= x^4 -36x^2$

We can find the first derivate and we got:

$y' = 4x^3 -72x$

In order to find the concavity we can find the second derivate and we got:

$y'' = 12x^2 -72$

We can set up this derivate equal to 0 and we got:

$y'' =12x^2 -72=0$

And solving we got:

$x=\pm \sqrt{\frac{72}{12}} =\pm \sqrt{6}$

We can find the sings of the second derivate on the following intervals:

$(-\infty$ Concave up

$x=-\sqrt{6}, y =-180$ inflection point

$(-\sqrt{6}$ Concave down

$x=\sqrt{6}, y=-180$ inflection point

$(\sqrt{6}$ Concave up

8 0

2 years ago