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3 years ago

In the complex number -12i. what is the real part?

Mathematics

1 answer:

coldgirl [10]3 years ago

8 0

Answer:

the real part is part of -12i is 0

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Write an expression for “the quotient for y and 8”

jonny [76]

Answer:

Step-by-step explanation:

I think your answer is y/8. It's a little unusually worded. If this is not correct, please leave a note.

The way I have answered it, I would word it as "What is the quotient of y divided by 8."

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3 years ago

Choose the improper fraction that is equivalent to the mixed number 2 17/24

Alborosie

Answer:

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Step-by-step explanation:

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2 years ago

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2 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

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2 years ago

In the complex number -12i. what is the real part?​

In the complex number -12i. what is the real part?