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3 years ago

14

I need help, algebra is hard

2 answers:

jasenka [17]3 years ago

5 0

Answer:

c

Step-by-step explanation:

aleksandr82 [10.1K]3 years ago

3 0

Answer: C

Step-by-step explanation:

I think this one is it?

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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

(x + 2) is raised to the fifth power. The third term of the expansion is:

morpeh [17]

Answer:

3

Step-by-step explanation:

6 0

3 years ago

-k=7<br>what is k equal to

kifflom [539]

-k=7

divide each side by -1

-k/-1 = 7/-1

k = -7

3 0

4 years ago

Read 2 more answers

Solve the system of linear equations and show your work.<br> −8x+5y=26<br> 9x−3y=−3

maks197457 [2]

X is 7 and y is 2 I am pretty sure

6 0

3 years ago

A high school has 44 players on the football team. The summary of the players' weights is given in the box plot. Approximately,

Marina CMI [18]

Answer:

Step-by-step explanation:

The missing image is attached below.

From the given information

The number of players is 44.

From the box plot;

The minimum weight = 154 pounds

The first quartile Q₁ = 159 which is 25% of the data below 159 pounds.

The second quartile Q₂ = 213 which is 50% of the data below 213 pounds

The third quartile Q₃ = 253 which is 75% of the data below 253 pounds

The maximum weight = 268 pounds.

Here, the value of 213 is the middle value and which signifies the median.

The 50% of the data are located both on the left side and the right side of the median.

Thus, the percentage of players weighting greater than or equal to 213 pounds is 50%.

7 0

3 years ago