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butalik [34]
2 years ago
11

5+3x write in verbal expression

Mathematics
2 answers:
natali 33 [55]2 years ago
6 0
Three times x plus five
Keith_Richards [23]2 years ago
5 0

Answer:

Five plus free equals 7

Step-by-step explanation:

I added five plus three

You might be interested in
EXERC
eimsori [14]

Answer:551.3212cm³

Step-by-step explanation:

Find the image attached

The volume is made up of a cone, cylinder and a hemisphere

Volume of the shape = Volume of cone + volume of cylinder + volume of hemisphere

Get the volume of the cone;

Volume of a cone Vc = 1/3πr²h

r is the base radius = 3.5cm

Height = 10cm

Vc = 1/3π(3.5)²(6)

Vc = 1/3π(12.25)(6)

Vc = 12.25 * 2π

Vc = 24.5π cm³

Get the volume of the cylinder;

Vcy = πr²h

r = 3.5cm

h = 10cm

Vcy = π(3.5)²(10)

Vcy = π(12.25)(10)

Vcy = π(122.5)

Vcy = 122.5π cm³

Get rhe volume of the hemisphere;

Volume of hemisphere = 2/3 πr³

r = 3.5cm

Vh =  2/3 π(3.5)³

Vh = 2/3π(42.875)

Vh = 28.58π cm³

Volume of the shape = VC + Vcy + Vh

Volume of the shape = 24.5π+122.5π+28.58π

Volume of the shape = 175.58π

<em>Volume of the shape = 551.3212cm³</em>

7 0
3 years ago
help asap im in a quiz, please show some sort of proof. and please dont say,¨ the answer is in the photo, and then send me a lin
Setler [38]
The answer would be D i think because if the two angles are to equal 72, and one of the angles equals 42, 72-42= 30. to find x, you’d do 30/6 which equals 5. so x=5 and the angle measure would be 30
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
How many times does nine go into 42
Lady bird [3.3K]
It only goes in 4 times
4 0
3 years ago
Pherris is graphing the function f(x) = 2(3)x. He begins with the point (1, 6). Which could be the next point on his graph?
Maurinko [17]

the next point would be (2,18)

4 0
3 years ago
Read 2 more answers
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