ANSWER ASAP Which statements about the values of 4.099 and 40.99 are true? Select all that apply.

Given that Ray E B bisects ∠CEA, which statements must be true? Select three options. m∠CEA = 90° m∠CEF = m∠CEA + m∠BEF m∠CEB =

FromTheMoon [43]

Answer:

Here is the question attached with.

$m\angle CEA =90 \ (deg)$

$m\angle BEF=135\ (deg)$

$\angle CEF$ is a straight line.

$\angle AEF$ is a right angled triangle.

Options $1,4,5,6$ are correct answers.

Step-by-step explanation:

⇒As $\ ray\ AE$ is ⊥ $FEC$ so it will forms right angled triangle then $m\angle CEA =90\ (deg)$ .

⇒Measure of $\angle BEF =135\ (deg)$ as $\angle BEF =\angle AEB +\angle AEF = (45+90)=135\ (deg)$ as $\angle AEB$ is the bisector of $\angle AEC$ ,meaning that $\angle AEB$ is half of $\angle AEC$ so $\angle AEB = 45\ (deg)$ .

⇒ $\angle CEF$ is a straight line as the angles measure over it is $180\ (deg)$ .

⇒Measure of $\angle AEF = 90\ (deg)$ from linear pair concept.

As $\angle CEA + \angle AEF = 180\ (deg)$ ,plugging the values of $m\angle CEA =90\ (deg)$ we have $\angle AEF = 90\ (deg)$ .

The other two options are false as:

$m\angle CEF=m\angle CEA + m\angle BEF = (90+135)=225$

it is exceeding $180\ (deg)$ whereas $\angle CEF$ is a

straight line.

And $m\angle CEB=2(m\angle CEA)$ is not true.

As $\angle CEA = 90\ (deg)$ and $\angle CEB=45\ (deg)$

So we have total $4$ answers.

The correct options are $1,4,5,6$ .

5 0

2 years ago

Read 2 more answers

3) You find a jar of quarters on the sidewalk and decide to start collecting them to cash in at the end of the school year.

soldier1979 [14.2K]

Answer:

I dont rally know

Step-by-step explanation:

try it yourself

5 0

1 year ago

One of the vertices of △PQR is P(2, −1). The midpoint of PQ is M(3, 0). The midpoint of QR is N(5, 3). Show that MN || PR and MN

VLD [36.1K]

Answer:

See the proof below

Step-by-step explanation:

Given the following coordinates

P(2, −1)

Midpoint of PQ M(3, 0)

We can get the coordinate point Q using the midpoint formula;

M(X,Y) = (x1+x2/2, y1+y2/2)

X = x1+x2/2

3 = 2+x2/2

6 = 2+x2

x2 = 6-2

x2 = 4

Y = y1+y2/2

0 = -1+y2/2

0 = -1 + y2

y2 = 0+1

y2 = 1

Hence the coordinate of Q is (4, 1)

Next is to get the coordinate of R

Given the midpoint of QR to be N(5, 3)

(5,3) = (4+x2/2, 1+y2/2)

5 = 4+x2/2

10 = 4+x2

x2 = 10-4

x2 = 6

1+y2/2 = 3

1+y2 = 6

y2 = 6-1

y2 = 5

Hence the coordinate of R is (6,5)

Given the coordinates M(3, 0) and N(5, 3)

Slope is expressed as:

m = y2-y1/x2-x1

m = 3-0/5-3

m = 3/2

Slope of MN = 3/2

Get the slope of PR

Given the coordinates P(2, −1) and R (6,5)

Slope of PR = 5-(-1)/6-2

Slope of PR = 5+1/4

Slope of PR = 6/4 = 3/2

Since the slope of MN is equal to that of PR, hence MN is parallel to PR i.e MN || PR

To show that MN = 1/2PR, we will have to take the distance between M and N and also P and R first as shown:

For MN with coordinates M(3, 0) and N(5, 3)

MN = √(x2-x1)²+(y2-y1)²

MN = √(5-3)²+(3-0)²

MN = √2²+3²

MN = √13

Get the length of PR where P(2, −1) and R (6,5)

PR = √(6-2)²+(5+1)²

PR = √4²+6²

PR = √16+36

PR = √52

PR = √4*13

PR = √4*√13

PR = 2√13

Since MN = √13

PR = 2MN

Divide both sides by 2

PR/2 = 2MN/2

PR/2 = MN

Hence MN = 1/2 PR (Proved!)

8 0

1 year ago

At the end of the first school day, Mr. Davies, a first grade teacher, inspected his classroom crayons. 2 of the 200 crayons w

MrRissso [65]

99% were not broken as only 1% were broken

4 0

2 years ago

Read 2 more answers

Bill launched a model rocket, and estimated its height h, In feet, after 1 seconds. His results are shown in the table

gizmo_the_mogwai [7]

Answer:

Bill launched a model rocket, and estimated its height h, In feet, after 1 seconds. His results are shown in the table

Time, 1 0 1 2 3 4

Height, h 0 110 190 240 255

Bill's data can be modeled by the function h(t) = -1612 + 128.

Which value is the best prediction for the height of the rocket after 5.5 seconds?

A 150 ft

B. 180 ft

C. 220 ft

D. 250 ft

E 260 ft

0 1

2 years ago