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3 years ago

Would appreciate some help first to answer gets brainliest

Mathematics

1 answer:

elixir [45]3 years ago

3 0

The correct answer is B

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What kind of function is 3y-2=x+7

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3 years ago

ABCD is a parallelogram. AB>AD. Prove: m∠ADB>m∠BDC. Show your work (statement reason maybe?)

Gwar [14]

In the parallelogram ABCD, join BD.

Consider the triangle Δ ABD.

It is given that AB > AD.

Since, in a triangle, angle opposite to longer side is larger, we have,

∠ ADB > ∠ ABD. --- (1)

Also, AB || DC and BD is a transversal.

Therefore,

∠ ABD = ∠ BDC

Substitute in (1), we get,

∠ ADB > ∠ BDC.

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Use mental math to solve the problem 8.02 − 5.98

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3 years ago

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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