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liraira [26]
3 years ago
7

Solve 19cos θ = 5 cos θ + 14 for all values of θ if θ is measured in radians.

Mathematics
1 answer:
Bas_tet [7]3 years ago
6 0

Answer:

Ф=xπ, x={0,+∞}

Step-by-step explanation:

be 19cosФ=5cosФ+14 → 19cosФ-5cosФ-14 → 14cosФ-14=0 → cosФ=1

then,  values for which cosФ=1 are Ф=0º=360º=720º........., but the values of Ф  must be radians, so Ф=xπ, where x∈Z, not including the negative numbers,

x={0,1,2,3,....∞}

Note: π≅3.141592

when calculating the values, it must be done in radian mode

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A source of information randomly generates symbols from a four letter alphabet {w, x, y, z }. The probability of each symbol is
koban [17]

The expected length of code for one encoded symbol is

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha\ell_\alpha

where p_\alpha is the probability of picking the letter \alpha, and \ell_\alpha is the length of code needed to encode \alpha. p_\alpha is given to us, and we have

\begin{cases}\ell_w=1\\\ell_x=2\\\ell_y=\ell_z=3\end{cases}

so that we expect a contribution of

\dfrac12+\dfrac24+\dfrac{2\cdot3}8=\dfrac{11}8=1.375

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By definition of variance, we have

\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2

For a string consisting of one letter, we have

\displaystyle\sum_{\alpha\in\{w,x,y,z\}}p_\alpha{\ell_\alpha}^2=\dfrac12+\dfrac{2^2}4+\dfrac{2\cdot3^2}8=\dfrac{15}4

so that the variance for the length such a string is

\dfrac{15}4-\left(\dfrac{11}8\right)^2=\dfrac{119}{64}\approx1.859

"squared" bits per encoded letter. For a string of length n, we would get \mathrm{Var}[L]=1.859n.

5 0
3 years ago
Which triangles could you use the HL Theorem to prove that Δ 1 = Δ 2?
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Number 3. I think so.
7 0
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puteri [66]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

Hope this helps :)

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bella types at a constant rate of 42 per minute. is the number of words she can type proportional to the number of minutes she t
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