Solve <br> Y/2 +22=38 for .<br><br> do u mind answering for some points?

Bingel [31]

Answer:

En álgebra, el teorema del factor es un teorema que vincula factores y ceros de un polinomio. Es un caso especial del teorema del resto polinómico.

Step-by-step explanation:

6 0

3 years ago

In Sakura's garden, for every 5 red flowers, there are 10 yellow flowers. There are 75 total red and yellow flowers. How many re

sweet [91]

Answer: there are 25 red flowers in Sakura's garden.

Step-by-step explanation:

In Sakura's garden, for every 5 red flowers, there are 10 yellow flowers. This means that if there are 10 red flowers, there would be 20 yellow flowers. if there are 20 red flowers, there would be 40 yellow flowers. Therefore, the ratio of red flowers to yellow flowers in the garden is

5/10 = 10/20 £= 20/40 = 1/2 = 1:2

Total ratio = 2 + 1 = 3

If there are 75 total red and yellow flowers, then the total number of red flowers are in Sakura's garden would be

1/3 × 75 = 25

8 0

3 years ago

What is the correct answer?

DochEvi [55]

The third one down is the correct answer. (-55 and 1)

4 0

3 years ago

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

Completely factor the polynomial below.

Blizzard [7]

The first step for solving this expression is to write 6q as a difference.
8q² + 10q - 4q - 5
Now factor out 2q from the first part of the expression (8q² + 10q).
2q × (4q + 5) - 4q - 5
Extract the negative sign from the second part of the expression (-4q - 5).
2q × (4q + 5) - (4q + 5)
Lastly,, factor out 4q + 5 from the expression to find your final answer.
(4q + 5) × (2q - 1)
This means that the correct answer to your question is (4q + 5) × (2q - 1),, or option D.
Let me know if you have any further questions.
:)

7 0

3 years ago

Solve Y/2 +22=38 for . do u mind answering for some points?