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3 years ago

PLEASE HELP!!!

Mathematics

2 answers:

damaskus [11]3 years ago

7 0

you would set the problem up and the answer will be 3 if you need me to explain it more comment down below

Anestetic [448]3 years ago

3 0

You would set the problem into slope-interslope form : y = mx + b.
Slope (m) = rate of change b = y-intercept
Therefore, your in the above:
m = 3
b = 0
The constant proportionality (slope) is 3 and the y-intercept is 0.

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What is the average rate of change for this function for the interval from x= 1

Usimov [2.4K]

Answer:

C. 8

Step-by-step explanation:

The average rate of change is the slope of the line between the two points of interest. It is given by the formula ...

m = (y2 -y1)/(x2 -x1)

For points (1, 2) and (3, 18), the average rate of change is ...

m = (18 -2)/(3 -1) = 16/2 = 8

The average rate of change between x=1 and x=3 is 8.

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2 years ago

Trey is solving his equation, and the last line of his work is "-2 = 2". What does this mean?

oksian1 [2.3K]

Answer:

it means that his equation is not true.

Step-by-step explanation:

-2=2 is obviously incorrect, so we must assume that the equation isn't true.

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3 years ago

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

Use the Distributive Property to solve the equation.

aliya0001 [1]

Answer:

x = -6

Step-by-step explanation:

-5(x + 3) = 15

Distribute the -5

-5x -15 =15

Add 15 to each side

-5x-15+15 =15+15

-5x = 30

Divide each side by -5

-5x/-5 = 30/-5

x =-6

8 0

3 years ago

Estimate -9 1/6 + 2 1/3 + (-1 7/8) I have more questions so if you want more just simply ask.

BigorU [14]

-8.70833333333 about but you can round to the nearest tenth (-8.7) or nearest whole number (-9)

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3 years ago