All categories

2 years ago

Find the midpoint of side EF. (4 points)

Mathematics

1 answer:

umka2103 [35]2 years ago

3 0

Answer:

Midpoint of side EF would be (-.5,4.5)

Step-by-step explanation:

We know that the coordinates of a mid-point C(e,f) of a line segment AB with vertices A(a,b) and B(c,d) is given by:

e=a+c/2,f=b+d/2

Here we have to find the mid-point of side EF.

E(-2,3) i.e. (a,b)=(2,3)

and F(1,6) i.e. (c,d)=(1,6)

Hence, the coordinate of midpoint of EF is:

e=-2+1/2, f=3+6/2

e=-1/2, f=9/2

e=.5, f=4.5

SO, the mid-point would be (-0.5,4.5)

You might be interested in

Which of the following is equivalent to –2i(6 – 7i)? (0 – 2i)(6 – 7i) (0 + 2i)(6 – 7i) (–2 + i)(6 – 7i) (6 – 2i) – (7i – 2i)

Mariulka [41]

Answer:

The expression which is equivalent to –2i(6 – 7i) is; (0 – 2i)(6 – 7i)

4 0

2 years ago

Check all the answers that represent the same level of precision,

Lostsunrise [7]

100 Is I was just like oh oh and then you are in the church

6 0

3 years ago

Can Someone Answer My Final Answers :) I’d appreciate it so much :)

DaniilM [7]

1. Remember that the perimeter is the sum of the lengths of the sides of a figure.To solve this, we are going to use the distance formula: $d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
where
$(x_{1},y_{1})$ are the coordinates of the first point
$(x_{2},y_{2})$ are the coordinates of the second point
Length of WZ:
We know form our graph that the coordinates of our first point, W, are (1,0) and the coordinates of the second point, Z, are (4,2). Using the distance formula:
$d_{WZ}= \sqrt{(4-1)^2+(2-0)^2}$
$d_{WZ}= \sqrt{(3)^2+(2)^2}$
$d_{WZ}= \sqrt{9+4}$
$d_{WZ}= \sqrt{13}$

We know that all the sides of a rhombus have the same length, so
$d_{YZ}= \sqrt{13}$
$d_{XY}= \sqrt{13}$
$d_{XW}= \sqrt{13}$

Now, we just need to add the four sides to get the perimeter of our rhombus:
$perimeter= \sqrt{13} + \sqrt{13} + \sqrt{13} + \sqrt{13}$
$perimeter=4 \sqrt{13}$
We can conclude that the perimeter of our rhombus is $4 \sqrt{13}$ square units.

2. To solve this, we are going to use the arc length formula: $s=r \alpha$
where
$s$ is the length of the arc.
$r$ is the radius of the circle.
$\alpha$ is the central angle in radians

We know form our problem that the length of arc PQ is $\frac{8}{3} \pi$ inches, so $s=\frac{8}{3} \pi$ , and we can infer from our picture that $r=15$ . Lest replace the values in our formula to find the central angle POQ:
$s=r \alpha$
$\frac{8}{3} \pi=15 \alpha$
$\alpha = \frac{\frac{8}{3} \pi}{15}$
$\alpha = \frac{8}{45} \pi$

Since $\alpha =POQ$ , We can conclude that the measure of the central angle POQ is $\frac{8}{45} \pi$

3. A cross section is the shape you get when you make a cut thought a 3 dimensional figure. A rectangular cross section is a cross section in the shape of a rectangle. To get a rectangular cross section of a particular 3 dimensional figure, you need to cut in an specific way. For example, a rectangular pyramid cut by a plane parallel to its base, will always give us a rectangular cross section.
We can conclude that the draw of our cross section is:

6 0

3 years ago

Read 2 more answers

In March Tom had -£165 in his bank account. In April he had -£156. In which month did he have the least amount of money in his a

Flura [38]

March = -165
April = -156

In march he is negative MORE money, meaning he has MORE money in April

This question is trying to confuse you, ussually the lowest number means less, but these numbers are negative, so the lowest number means more.
He has the least amount of money in March

5 0

2 years ago

Question #2: Pythagorean Theorem

astra-53 [7]

Answer:

isnfdytsafksjd

Step-by-step explanation:

ryref w4trfd ws 3wqrs

3 0

2 years ago