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BartSMP [9]
2 years ago
5

Help me . With work to plz

Mathematics
1 answer:
Zepler [3.9K]2 years ago
7 0

Answer:

57°

Step-by-step explanation:

Angles EGB and EHD are corresponding angles of two parallel lines cut by a transversal; therefore, they are congruent.

m<EGB = m<EHD = 57°

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y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence
sukhopar [10]

Let

\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots

Differentiating twice gives

\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots

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

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

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

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

Then the coefficients in the power series solution are governed by the recurrence relation,

\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

k=0 \implies n=0 \implies a_0 = a_0

k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}

k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}

k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}

It should be easy enough to see that

a_{n=2k} = \dfrac{a_0}{(2k)!}

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

k = 0 \implies n=1 \implies a_1 = a_1

k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}

k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}

k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}

so that

a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}

So, the overall series solution is

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

\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}

4 0
2 years ago
4.5 rounded to the nearest hole number
Snowcat [4.5K]

Answer: 5

Step-by-step explanation:

5 0
2 years ago
How to solve 2x(x+3)=2x^2+15
Travka [436]

Answer: x = 5/2 = 2.5

Step-by-step explanation:

<u>Given expression</u>

2x (x + 3) = 2x² + 15

<u>Expand parentheses and apply the distributive property</u>

2x · x + 2x · 3 = 2x² + 15

2x² + 6x = 2x² + 15

<u>Subtract 2x² on both sides</u>

2x² + 6x - 2x² = 2x² + 15 - 2x²

6x = 15

<u>Divide 6 on both sides</u>

6x / 6 = 15 / 6

x = 15/6

\boxed{x=\frac{5}{2} }

Hope this helps!! :)

Please let me know if you have any questions

5 0
3 years ago
Read 2 more answers
4 cot theta<br>=<br>tan theta<br>how do I I find the value of quadrant?​
Ray Of Light [21]

Answer:

4cotα=tanα

4(1/tanα)=tanα

(4/tanα)=tanα

cross multiply

=> 4=tan²α

√4=√tan²α

±2=tanα

α=arc( tan) |2|

α=63.4° ( in first quadrant)

and

α=180+63.4=243.4 in the third quadrant

since we also found a negative answer( i.e –2) then α also lies in quadrants where it gives a negative value(i.e second and fourth quadrants)

α=180–63.4=116.6° in the second quadrant

α=360–63.4=296.6 in the fourth quadrant

therefore theta( in my case, alpha) lies in all four quadrants and is equal to:

α=63.4°,243.4°,116.6°and 296.6°

5 0
3 years ago
A 2000 lb crystal chandelier in a Las vegas Hotel is suspended by 2 cables.
Musya8 [376]

Answer:

maaf tidak tau ya saya wkwkwk

5 0
3 years ago
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