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goblinko [34]
3 years ago
9

Compute the sum of the squares of the shorter lengths. Compute the sum of the squares of the shorter lengths. Compute the square

of the longest length. Compute the square of the longest length. What kind of triangle is it? What kind of triangle is it? Acute triangle Acute triangle Right triangle Right triangle Obtuse triangle Obtuse triangle
Mathematics
1 answer:
fomenos3 years ago
6 0

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete as the sides of the triangle are not given.

However, I'll answer this question using a general scope.

Let the sides of the triangle be w, x and y.

Where w and x are the shorter sides.

The following conditions exist:

1. If w² + x² < y², then the triangle is obtuse.

2. If w² + x² > y², then the triangle is acute

3. Lastly, if w² + x² = y², then the triangle is right angled

So, take the sides of the triangles; take square of the two shorter sides and compare it with the square of the longest side.

Compare this with the (3) conditions above to answer the question.

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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
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Let

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

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\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

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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

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\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}

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

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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)!}

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

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

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

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

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\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)!}}

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