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3 years ago

Trigonometry Problem Find the length of the missing side.

Mathematics

1 answer:

spin [16.1K]3 years ago

5 0

Answer:

19.0

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relationship ...

Cos = Adjacent/Hypotenuse

cos(38°) = 15/w

w = 15/cos(38°) ≈ 19.035

The length of the missing side is about 19.0.

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2 years ago

A hemispherical bowl of radius 5 inches is filled to a depth of h inches, where 0less than or equalshless than or equals5. Find

mestny [16]

The shape of the water is that of a "spherical cap." The formula for the volume of a spherical cap is ...

... V = π·h²·(r -h/3) . . . . . . . from web search

For a radius of 5 in, this is

... V = π·h²·(5 -h/3) . . . . in³

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For h=0

... V = π·0²·(5 -0/3) = 0

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3 years ago

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

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

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

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3 years ago

Would you solve the equation 0.25x + 7 = 1/3x – 8 <br> using fractions or decimals? Explain please.

mamaluj [8]

Answer:

$x = 180$

Step-by-step explanation:

Solve the value of $x$ :

$0.25x + 7 = \frac{1}{3}x - 8$

-Combine $\frac{1}{3}x$ and $0.25x$ by subtracting $0.25x$ by $\frac{1}{3}x$ :

$0.25x - \frac{1}{3}x + 7 = \frac{1}{3}x - \frac{1}{3}x - 8$

$-\frac{1}{12}x + 7 = -8$

-Subtract $7$ on both sides:

$-\frac{1}{12}x + 7 - 7 = -8 - 7$

$-\frac{1}{12}x = -15$

-Multiply both sides by $-12$ , the reciprocal of $-\frac{1}{12}$ :

$\frac{-\frac{1}{12}x}{-\frac{1}{12}} = \frac{-15}{-\frac{1}{12}}$

$x = -15 (-12)$

$x = 180$

Therefore, the value of $x$ is $180$ .

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3 years ago

Dimitri rode his bike 32 miles yesterday. He rode 12 4/5 miles before lunch and the rest of the distance after lunch. How far di

BigorU [14]

19.2 miles is the answer

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3 years ago

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