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jenyasd209 [6]
3 years ago
8

Complete the paragraph proof. We are given that Ray E B bisects ∠AEC. From the diagram, ∠CED is a right angle, which measures de

grees. Since the measure of a straight angle is 180°, the measure of angle must also be 90° by the . A bisector cuts the angle measure in half. m∠AEB is 45°.

Mathematics
2 answers:
Vilka [71]3 years ago
7 0

Answer: 1. 90, 2. AEC, 3. angle addition postulate

Step-by-step explanation:

Oduvanchick [21]3 years ago
4 0

Answer:

1. 90

2. AEC

3. angle addition postulate

:)

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9 x 7 = ( x 7) – (1 x 7)<br> = 70-<br> =
WARRIOR [948]

Answer:

63

Step-by-step explanation:

9x7=(10x7)-1x(7)

=70-7

=63

8 0
2 years ago
HELP FAST FAST FAST
Rus_ich [418]

Answer:

h=3in  

Step-by-step explanation:

A=a+b

2h

Solving forh

h=2A

a+b=2·13.5

3+6=3in

3 0
3 years ago
The poster shows how many of its games of football team has won so far.Express this information as a fraction a,percent and as a
Harrizon [31]
12/15
Divide 12 by 15 and then you should get 0.8
0.8 is obviously equal to 8/10ths or 80% 
Hope this helps :)
4 0
3 years ago
Can you help me please?
9966 [12]
First, change the two mixed numbers in the expression into an improper fraction: 25/6 + 5/3.
Then, find the Lowest Common Denominator of both fractions, which is 6, and set both denominators equal to that.  Remember, whatever you do on one side you must do to the other: 25/6 + 10/6
Add the two together: 25/6 +10/6 = 35/6.
To make it a mixed number again, find how many times 6 goes into 35, which is 5 times, with a remainder of 5.  Your answer is 5 5/6
6 0
3 years ago
What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po
emmasim [6.3K]
\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA
\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k
\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j

The cross product is

\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j

So, the flux is given by

\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA
\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du
\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv
\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv

where t=-\sin v in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi
5 0
3 years ago
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