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mr Goodwill [35]
2 years ago
14

Identify the constants from this expression. 5x - 7x + 5 - 12 + 4x - 3

Mathematics
1 answer:
skelet666 [1.2K]2 years ago
5 0

Answer:

5, -12, -3

Step-by-step explanation:

A constant is a number by itself. Here, you can see that 5X is paired with a letter, that means that 5X is not a constant. So you could say that any whole number is a constant number and in this problem, 5, -12 and -3 are all the constant numbers in this expression.

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5. Find the value of x that makes BC || DE.
s344n2d4d5 [400]

Answer:

x = -7

Step-by-step explanation:

To find the value of x, you need to set up a proportion:

\frac{x-2}{15} = \frac{x-1}{18}

Then cross multiply:

15x - 15 = 18x + 36

Then subtract 15x from both sides:

-15 = 3x + 36

Subtract 36 from both sides:

-51 = 3x

Finally, divide both sides by 3:

x = -7

8 0
3 years ago
WILL GIVE BRAINLIEST FOR CORRECT ANSWER
hammer [34]
Tan(35)=x/12
.7=x/12
12*.7=x
x=8.4

6 0
3 years ago
Read 2 more answers
A bus eats 4 people in a row and there are 12 rows how would fit on 2 buses?
BARSIC [14]

Answer:

4 \times 12 \times 2 = 96

96 people in total.

BTW its seats and I think you meant "how many"

6 0
3 years ago
URGENT HELP THIS IS FOR DWREAD ONLY
QveST [7]

Answer:

(-1,0) and (3,0)

Step-by-step explanation:

The x-intercepts are the points where the graph crosses the x-axis. (-1,0) and (3,0)

6 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
2 years ago
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