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3 years ago

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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3 years ago

Geometry help #31 pls

Anestetic [448]

Plug the coordinates into the distance formula and solve.

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3 years ago

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

melomori [17]

Step-by-step explanation:

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3 years ago

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

Step-by-step explanation: You have to make the 5 into a negative number using parenthesis and do the math.

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3 years ago

Read 2 more answers

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 $\mathrm{Var}[L]=1.859n$ .

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3 years ago