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3 years ago

Find the area of the shaded region in terms of x.

Mathematics

1 answer:

Tju [1.3M]3 years ago

4 0

Answer:

35x^2 - 71x + 15

Step-by-step explanation:

look at the photo

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5x+20y=4 x=-4y+1 Solve for x/y

Andrews [41]

Hi there!

Looks like a lot of work is going on here.

Let's start by solving this equation for the x.

$5x + 20y = 4$

Step 1) Add -20y to both sides.

$5x + 20y + - 20y = 4 - 20y 5x = -20y + 4$

Step 2) Divide both sides by 5.

$\frac{5x}{5} = \frac{-20y + 4}{5}$

$x = -4y + \frac{4}{5}$

Final Answer:

$x = -4y + \frac{4}{5}$

Now, let's solve for y.

Step 1) Add -5x to both sides.

$5x + 20y + - 5x = 4 + - 5x 20y = -5x + 4$

Step 2) Divide both sides by 20.

$\frac{20y}{20} = \frac{-5x + 4}{20}$

$y = \frac{-1}{4} x + \frac{1}{5}$

Final Answer:

$y = \frac{-1}{4} x + \frac{1}{5}$

Hope this helps! :D

Hint - When you have two variables in an equation, and you want to solve for one, the other will end up in your answer. :)

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

3 years ago

System of equations, special case infinitely many solutions?????

Mashcka [7]

One possible system is

1x + 3y = 4

2x + 6y = 8

Note how 2 is twice as large as 1, 6 is twice as large as 3, and 8 is twice as large as 4.

In other words, the second equation is the result of multiplying both sides of the first equation by 2.

1x+3y = 4

2*(1x+3y) = 2*4

2x+6y = 8

Effectively the two equations in bold are the same which produces the same line. The two lines overlap perfectly to intersect infinitely many times. An intersection is a solution.

4 0

3 years ago

How many matches are played together in a knockout tournament involving 36 teams?

Svetradugi [14.3K]

Answer:

need points srry

Step-by-step explanation:

8 0

3 years ago

What is the number of coefficients in 6a-3, 0.2-y+8z, and 1/2 n? Someone please help

hjlf

There are 3 coefficents. There is one in 6a, which is 6, one in 8z, which is 8, and one in 1/2n, which is 1/2.

Hope this helps, Ngam123!