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loris [4]
3 years ago
13

CAN SOMEONE PLEASE HELP ME WITH THIS?

Mathematics
2 answers:
Anarel [89]3 years ago
4 0

Answer:

B. ∠EBC and ∠ABE

D. ∠CBE and ∠ABD

C. ∠ABD and ∠ABE

Supplementary Angles are angles that add up to 180°

In this specific shape, angles that equal 180° are the angles that go across the whole line, NOT the vertical angles, like your last question. So picture it like this: The only angles that are supplementary should go from:

Point D to E

Point A to C

Hopefully this helped you!

Ad libitum [116K]3 years ago
3 0

Answer:

EBC and ABE

ABD and ABE

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vlada-n [284]
It can be any number so here is my awnser 437
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4 years ago
Last qestion if answered 40 points and brianlist and if right
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Answer:

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Step-by-step explanation:

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6 0
3 years ago
1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING
iogann1982 [59]

If

y=\displaystyle\sum_{n=0}^\infty a_nx^n

then

y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n

The ODE in terms of these series is

\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0

\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0

\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}

We can solve the recurrence exactly by substitution:

a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0

\implies a_n=\dfrac{(-2)^n}{n!}a_0

So the ODE has solution

y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}

which you may recognize as the power series of the exponential function. Then

\boxed{y(x)=a_0e^{-2x}}

7 0
3 years ago
Find a formula for the distance from the point​ P(x,y,z) to each of the following planes. a. Find the distance from​ P(x,y,z) to
BARSIC [14]

Answer:

Distance to the xy-plane = |z|

Distance to the yz-plane = |x|

Distance to the xz-plane = |y|

Step-by-step explanation:

The distance from P(x,y,z) to the xy-plane is by definition the magnitude of the vector that goes from the perpendicular projection of P over the xy-plane to the point P, which is exactly the magnitude of the vector (0,0,z) = |z| the absolute value of z

Similarly, the distance from P to the yz-plane is |x| and the distance from P to the xz-plane is |y|

Distance to the xy-plane = |z|

Distance to the yz-plane = |x|

Distance to the xz-plane = |y|

7 0
3 years ago
Are the ratios 6:2 and 2:6 equivalent?
Alenkasestr [34]

Answer:

no

Step-by-step explanation:

5 0
3 years ago
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