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Gnom [1K]
3 years ago
12

Choose the expression that represents the prime factorization of 96.

Mathematics
1 answer:
fredd [130]3 years ago
4 0

Answer:

3.2.2.2.2...

Step-by-step explanation:

The prime factors of 96 are written as 2 x 2 x 2 x 2 x 2 x 3 or 3 x 25, where 2 and 3 are the prime numbers

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Let the (x; y) coordinates represent locations on the ground. The height h of
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The critical points of <em>h(x,y)</em> occur wherever its partial derivatives h_x and h_y vanish simultaneously. We have

h_x = 8-4y-8x = 0 \implies y=2-2x \\\\ h_y = 10-4x-12y^2 = 0 \implies 2x+6y^2=5

Substitute <em>y</em> in the second equation and solve for <em>x</em>, then for <em>y</em> :

2x+6(2-2x)^2=5 \\\\ 24x^2-46x+19=0 \\\\ \implies x=\dfrac{23\pm\sqrt{73}}{24}\text{ and }y=\dfrac{1\mp\sqrt{73}}{12}

This is to say there are two critical points,

(x,y)=\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\text{ and }(x,y)=\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)

To classify these critical points, we carry out the second partial derivative test. <em>h(x,y)</em> has Hessian

H(x,y) = \begin{bmatrix}h_{xx}&h_{xy}\\h_{yx}&h_{yy}\end{bmatrix} = \begin{bmatrix}-8&-4\\-4&-24y\end{bmatrix}

whose determinant is 192y-16. Now,

• if the Hessian determinant is negative at a given critical point, then you have a saddle point

• if both the determinant and h_{xx} are positive at the point, then it's a local minimum

• if the determinant is positive and h_{xx} is negative, then it's a local maximum

• otherwise the test fails

We have

\det\left(H\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)\right) = -16\sqrt{73} < 0

while

\det\left(H\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)\right) = 16\sqrt{73}>0 \\\\ \text{ and } \\\\ h_{xx}\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-8 < 0

So, we end up with

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