Please help me plz im begging u

What is 2-2? i dont know

12345 [234]

I think 0? Correct me if I’m wrong

5 0

3 years ago

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Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago

Solve for 3.3x-4=-13.9

evablogger [386]

Answer: Simple and best practice solution for 3.3x-4=-13.9 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.

Step-by-step explanation: Idk how to explain I’m just smart so I can figure it out

4 0

3 years ago

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Assume the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons.

Rama09 [41]

The rate of change is 16.8 million tons per year

Solution:

Given that, the world catch of fish in 1950 was 12 million tons and in 1955 it was 96 million tons

The average rate of change is given by formula:

$\text{ Average rate of change } = \frac{\text{ change in value }}{\text{ number of years }}$

Value in 1950 = 12 million tons

Value in 1955 = 96 million tons

Change in value = value in 1955 - value in 1950

Change in value = 96 million - 12 million = 84 million tons

Number of years = 1950 to 1955 = 5 years

Substitute the given values in formula,

$\text{ Average rate of change } = \frac{84 \text{ million tons}}{5} = 16.8 \text{ million tons}$

Thus rate of change is 16.8 million tons per year

5 0

3 years ago

The answer and explanation to question 1

xxMikexx [17]

Answer:

20, 31, 33, 35, 55, 57, 58, 59, 72, 73, 79, 86, 87, 88

Step-by-step explanation:

The data set is split by digit. So 2|0 becomes 20. 5|5 becomes 55. Using this method, the data set is

20, 31, 33, 35, 55, 57, 58, 59, 72, 73, 79, 86, 87, 88

7 0

3 years ago