All categories

3 years ago

I need help please?!!!

Mathematics

1 answer:

san4es73 [151]3 years ago

8 0

Answer: (3)/(4)(3x+8), x^(2)-4, 3^(2+)x, -5(x-3)

Step-by-step explanation: Those are the right answers, trust me, this took a while.

Have a good day, brainliest wouldn't be bad :)

You might be interested in

Three more than the product of two and a number

Sati [7]

Answer:

2x+3

Step-by-step explanation:

+3 " three more than"

2x " the product of two and a number"

2x+3 " three more than the product of two and a number"

8 0

3 years ago

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

What quadrilateral always has 4 congruent angles and opposite sides that are congruent and parellel

Nata [24]

I’m pretty sure it’s a Rectangle or square

5 0

3 years ago

How do you solve this?

snow_tiger [21]

Answer:

-368

Step-by-step explanation:

5 0

4 years ago

On a December day, the probability of snow is 0.30. The probability of a "cold" day is 0.50. The probability of snow and "cold"

Margarita [4]

Answer:

Yes, snow and cold weather are independent.

Step-by-step explanation:

We are given the following in the question:

C: Cold weather

S: Snow

P(C) = 0.50

P(S) =0.30

$P(S\cap C) = 0.15$

We have to check whether snow and cold whether are independent events.

If the events A and B are independent then,

$p(A\cap B) = P(A)\times P(B)$

Checking,

$p(S\cap C) = P(S)\times P(C)\\0.15 = 0.30\times 0.50$

Thus, the two events are independent.

For mutually exclusive events

$P(A\cap B) = 0$

Thus, the given events are not mutually exclusive.

8 0

3 years ago