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Lady_Fox [76]
3 years ago
15

What is the angle of GXI

Mathematics
1 answer:
bekas [8.4K]3 years ago
7 0

Answer:

it is 22

Step-by-step explanation:

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How is “the product of 208 anda is 256” written as a equation?
BigorU [14]

Answer: 208a = 256

Step-by-step explanation:

product means multiplication so

208a = 256

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Pls help im giving brainliest if you can help me
vovangra [49]

Answer:

Here it is!!

Step-by-step explanation:

It is the same as you will encounter the same things as u would while adding integers. Negative plus negative would equal negative. etc. It would be different because you need to take the numbers on the right side of the decimal into account. Basically, -2.66 + -2.66 = -5.32

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Which of the following is equivalent to 9^5/2.
Korvikt [17]

Answer:

a

Step-by-step explanation:

3 0
3 years ago
Factor completely 2x2 – 32.
trapecia [35]

Answer: -28

Step-by-step explanation:

4-32= -28

8 0
3 years ago
Let the (x; y) coordinates represent locations on the ground. The height h of
grigory [225]

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

h\left(\dfrac{23+\sqrt{73}}{24},\dfrac{1-\sqrt{73}}{12}\right)=-\dfrac{4247+37\sqrt{73}}{72} \text{ (saddle point)}\\\\\text{ and }\\\\h\left(\dfrac{23-\sqrt{73}}{24},\dfrac{1+\sqrt{73}}{12}\right)=-\dfrac{4247-37\sqrt{73}}{72} \text{ (local max)}

7 0
3 years ago
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