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almond37 [142]
3 years ago
15

Someone pls help me I need help plsssss

Mathematics
1 answer:
Lyrx [107]3 years ago
3 0

Answer:

3(x+1)=2(x-1)

3x+3=2x-2

3x-2x=-2-3

x=-5.

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What geometric term is named by its end point
GuDViN [60]

Answer: A line segment represents a collection of points inside the endpoints and it is named by its endpoints.

Also, These words are point, line and plane, and are referred to as the three undefined terms of geometry.

* Hopefully this helps:) Mark me the brainliest:)!!!

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
Math class, need help on this. Anybody?? ​
ASHA 777 [7]

Answer:

y = 2x-5

Step-by-step explanation:

y = 2x+2

We know the slope of this line is 2 because it is in the form y =mx +b  where m is the slope

We want a parallel line so it has the same slope

We can use the point slope form of the equation

y-y1 = m(x-x1)

y--1 = 2(x-2)

y+1 = 2(x-2)

Distribute

y+1 = 2x-4

Subtract 1 from each side

y+1-1 = 2x-4-1

y = 2x-5

This is in slope intercept form

3 0
3 years ago
Mary says,
igor_vitrenko [27]

Answer: you could look it up or have a tutuar teach you to learn it

Step-by-step explanation:

7 0
3 years ago
1) What would be another way to name plane M?
garri49 [273]

The correct answer is C plane HJG

4 0
3 years ago
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