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3 years ago

7

An equation for the line graphed is?

2 answers:

alex41 [277]3 years ago

8 0

To solve this problem, we can begin by analyzing the graph and then eliminating incorrect answers. By looking at the graph, we can see that it has a positive correlation (the graph is increasing from left to right), which tells us that it has a positive slope. Using this knowledge, we can eliminate answer choices C and D, which both suggest equations with a negative slope (which would mean that the line was decreasing from left to right).

Next, we must analyze the slope. Answer A suggests that the slope is 3/2, meaning that the line rises 3 units for every 2 units it goes across. Answer B suggests that the slope is 1/2, meaning that the line rises 1 unit for every 2 units that it goes across. If we look at the graph, let's say at the point (0,3), the next closest point is 3 units up and 2 units across, which lets us know that the slope is positive 3/2.

Therefore, your answer is A. y = 3/2x + 3 (which is confirmed by the point we recognized, (0,3), which is the y-intercept of the equation).

Hope this helps!

liubo4ka [24]3 years ago

4 0

I would say the answer is : A. y=3/2x +3

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Find the value of x, if angle x measures 102 degrees

Over [174]

The answer is B
Have a good day!

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3 years ago

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The distance between the two points pictured is d= Vn

katrin [286]

the distance between points is:

d = 7.8 units

d = root ((x2-x1) ^ 2 + (y2-y1) ^ 2)

The ordered pairs are:

(x1, y1) = (- 3, -2)

(x2, y2) = (2,4)

By applying the formula we have:

d = root ((2 - (- 3)) ^ 2 + (4 - (- 2)) ^ 2)

d = root (61)

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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

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blagie [28]

it's (0, 0) and (-2, 0) so A & C

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Can anyone help me with 3 and 4

Gnoma [55]

No as they do not simplify into the same fraction
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2 years ago