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Anni [7]
3 years ago
6

Which algebraic expression is a trinomial?

Mathematics
1 answer:
Goryan [66]3 years ago
7 0

Answer: A trinomial consisting of three terms.

Step-by-step explanation: An example of trinomial is 6x squared + 3x + 5.

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Find the length of a side of a square if its area is:<br> 0.49 square units
barxatty [35]

Answer:

Side = 0.7 units

Step-by-step explanation:

Area of a square = 0.49 sq. units

Side² = 0.49

side² = \frac{49}{100}\\\\

side = \sqrt{\frac{49}{100}}=\sqrt{\frac{7*7}{10*10}}\\\\sides=\frac{7}{10}=0.7

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Plug the coordinates into the distance formula and solve.
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The point (3,-7) lies on a circle and the center of the circle is at (-2,-7). What is the equation of this circle
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the answer is in the above image

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Which of the following expressions shows how to rewrite 4 - 5 using the addtive inverse and displays the expression correctly on
Evgesh-ka [11]

Answer: 4 - 5 = 4 + (-5) = -1

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3 years ago
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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

bits to the code per encoded letter. For a string of length n, we would then expect E[L]=1.375n.

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
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