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4 years ago

12

Which choice does NOT match the shape with its correct number of sides?

1 answer:

Usimov [2.4K]4 years ago

7 0

The answer is heptagon because a heptagon is a shape with 7 sides, not 6. :)

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How many ordered quadruples (w, x, y, z) of positive odd integers are there

guajiro [1.7K]

Answer:

165

Step-by-step explanation:

6 0

3 years ago

Freeeeeeee POINTS!!!!!!! !!!!❤!!!!!

pychu [463]

Answer:

thanks i guess

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

One line passes through the points \blueD{(-3,-1)}(−3,−1)start color #11accd, (, minus, 3, comma, minus, 1, ), end color #11accd

mart [117]

Answer:

The lines are perpendicular

Step-by-step explanation:

we know that

If two lines are parallel, then their slopes are the same

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

Remember that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

<em>Find the slope of the first line</em>

we have the points

(-3,-1) and (1,-9)

substitute in the formula

$m_1=\frac{-9+1}{1+3}$

$m_1=\frac{-8}{4}$

$m_1=-2$

<em>Find the slope of the second line</em>

we have the points

(1,4) and (5,6)

substitute in the formula

$m_2=\frac{6-4}{5-1}$

$m_2=\frac{2}{4}$

Simplify

$m_2=\frac{1}{2}$

<em>Compare the slopes</em>

$m_1=-2$

$m_2=\frac{1}{2}$

Find out the product

$m_1*m_2=(-2)(\frac{1}{2})=-1$

therefore

The lines are perpendicular

8 0

3 years ago

Read 2 more answers

Find the critical points of the surface f(x, y) = x3 - 6xy + y3 and determine their nature.

Vedmedyk [2.9K]

Compute the gradient of $f$ .

$\nabla f(x,y) = \left\langle 3x^2 - 6y, -6x + 3y^2\right\rangle$

Set this equal to the zero vector and solve for the critical points.

$3x^2-6y = 0 \implies x^2 = 2y$

$-6x+3y^2=0 \implies y^2 = 2x \implies y = \pm\sqrt{2x}$

$\implies x^2 = \pm2\sqrt{2x}$

$\implies x^4 = 8x$

$\implies x^4 - 8x = 0$

$\implies x (x-2) (x^2 + 2x + 4) = 0$

$\implies x = 0 \text{ or } x-2 = 0 \text{ or } x^2 + 2x + 4 = 0$

$\implies x = 0 \text{ or } x = 2 \text{ or } (x+1)^2 + 3 = 0$

The last case has no real solution, so we can ignore it.

Now,

$x=0 \implies 0^2 = 2y \implies y=0$

$x=2 \implies 2^2 = 2y \implies y=2$

so we have two critical points (0, 0) and (2, 2).

Compute the Hessian matrix (i.e. Jacobian of the gradient).

$H(x,y) = \begin{bmatrix} 6x & -6 \\ -6 & 6y \end{bmatrix}$

Check the sign of the determinant of the Hessian at each of the critical points.

$\det H(0,0) = \begin{vmatrix} 0 & -6 \\ -6 & 0 \end{vmatrix} = -36 < 0$

which indicates a saddle point at (0, 0);

$\det H(2,2) = \begin{vmatrix} 12 & -6 \\ -6 & 12 \end{vmatrix} = 108 > 0$

We also have $f_{xx}(2,2) = 12 > 0$ , which together indicate a local minimum at (2, 2).

3 0

2 years ago

Explain how a number line can be used to find the difference for 34-28.

den301095 [7]

You can use a number line to find the difference of 34 - 28 by starting at 34 and go backwards 20 numbers, then 8 numbers.

8 0

3 years ago