Estimate of 137,638 and 52,091

Point C is in the interior of angle DEF. If angle DEC and angle CEF are congruent, then segment EC is the _____ of angle DEF.

emmasim [6.3K]

Answer:

Option A

Step-by-step explanation:

Line EC splits the angle into two equal angles so it is an angular bisector of DEF.

7 0

3 years ago

The graph of a linear equation passes through three of these points: (0, 5), (2, 2), (3, 1), and (4, -1). The point that the gra

dlinn [17]

Label the given points as follows:
A (0, 5)
B (2, 2)
C (3, 1)
D (4, -1)

The straight line has a constant slope. Therefore it should have the same value when any two of the four points are used to calculate the slope.

Try A and B:
Slope = (2 - 5)/(2 - 0) = -3/2

Try A and C:
Slope = (1 - 5)/(3 - 0) = -4/3

Try A and D.
Slope = (-1 - 5)/(4 - 0) = -3/2

Try B and D.
Slope = (-1 - 2)/(4 - 2) = -3/2

Clearly, C does not lie on the straight line.

Answer: The point that the graph does not pass through is (3,1).

5 0

3 years ago

Find the solution of the system of the equations 7x+14y=35 -2x-2y=-8

grandymaker [24]

Answer:

(x,y)=(309/7,-317/14) If you ever get stuck on similar problems I recommend using the photomath app.

Step-by-step explanation:

4 0

2 years ago

4 ^-3 rewrite without an exponent

hichkok12 [17]

Answer:

(1/4)³

Step-by-step explanation:

Hope it helps you

5 0

2 years ago

Read 2 more answers

Need help in number 12 and 13 PLEASEE!! I don’t get it and I start school the day after tomorrow

Natasha_Volkova [10]

Answer:

The explanations for the graphs are provided down below. Please let me know if you have any questions about my answer.

12 and 13 as written on the worksheet is right.

Step-by-step explanation:

12) The answer given is correct.

The relation between x and y is given as:

$y=\frac{x^2}{2}-3$ with $x \in \{-4,-2,0,2}$ .

I replaced the word domain with x since the domain is the set of x's for which the relation exists.

We are going to replace x with each of the x's given to see what y corresponds to each.

Let's begin with x=-4.

$y=\frac{x^2}{2}-3$ with $x=-4$ :

$y=\frac{(-4)^2}{2}-3$

$y=\frac{16}{2}-3$

$y=8-3$

$y=5$ .

So (-4,5) is an ordered pair that should be on our graph.

To find this point you move left 4 from origin then up 5. Now you put a dot where you have landed. Your graph does show this point.

Moving on.

Let's do the next x: x=-2.

$y=\frac{x^2}{2}-3$ with $x=-2$ :

$y=\frac{(-2)^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (-2,-1) is an ordered pair that should be on our graph.

To find this point you move left 2 from origin and then down 1. Now you put a dot where you have landed. Your graph shows this point as well.

Now x=0.

$y=\frac{x^2}{2}-3$ with $x=0$ :

$y=\frac{0^2}{2}-3$

$y=\frac{0}{2}-3$

$y=0-3$

$y=-3$

So (0,-3) is an ordered pair that should be on our graph.

To find this point you move left and right none and down 3. Now you put a dot where you have landed. Your graph shows this point.

Now the last point will be at x=2.

$y=\frac{x^2}{2}-3$ with $x=2$

$y=\frac{2^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (2,-1) is an ordered pair that should be on our graph.

To find this point you move 2 units right from the origin and then down 1 unit. Now put a dot where you landed. The graph shows this point as well.

13) The answer given is correct.

$g(x)=|x|$ is the parent function and makes like a V shaped graph where it's vertex is at (0,0).

If we want to move this graph right 3 it becomes:

$m(x)=|x-3| \text{ or } m(x)=|(-1)(-x+3)|=|-1||-x+3|=1|-x+3|=|-x+3|=|3-x|$ .

If you move that up once it becomes:

$n(x)=|x-3|+1$ or $n(x)=|3-x|+1$ which is the curve given.

If you don't know about transformations you can choose a few points to plug in to see what's going on with the graph.

Let's choose x=-5,-3,-1,0,1,3,5.

x=-5

f(-5)=|3--5|+1=|3+5|+1=|8|+1=8+1=9.

There is no room for (-5,9) on our graph but if you extended the left hand side of the absolute value function there you would see that (-5,9) is reached.

x=-3

f(-3)=|3--3|+1=|3+3|+1=|6|+1=6+1=7.

(-3,7) should be a point on the graph. Same thing for this point as (-5,9).

x=-1

f(-1)=|3--1|+1=|3+1|+1=|4|+1=4+1=5.

(-1,5) is located on the graph.

x=0

f(0)=|3-0|+1=|3|+1=3+1=4.

(0,4) is also located on the graph.

x=1

f(1)=|3-1|+1=|2|+1=2+1=3.

(1,3) is located on the graph.

x=3

f(3)=|3-3|+1=|0|+1=0+1=1.

(3,1) is located on the graph.

x=5

f(5)=|3-5|+1=|-2|+1=2+1=3.

(5,3) is located on the graph.

Now if we weren't given the graph already:

I would plot the points I found which were:

(-5,9)

(-3,7)

(-1,5)

(0,4)

(1,3)

(3,1)

(5,3)

We should get a basic idea of what the function looks like from these points.

I will graph them. You will have to connect these points because the domain isn't discrete like number 12 is. That is they didn't list out elements for the domain.

I'm going to graph one more point after x=5.

How about x=7?

$f(7)=|3-7|+1=|-4|+1=4+1=5$

So (7,5) is also a point on the graph.

You should see that the blue points are following the red path I made there.

8 0

2 years ago