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Licemer1 [7]
3 years ago
13

Use the roster method to write each of the given sets. For some exercises you may need to consult a reference, such as the Inter

net or an encyclopedia. (Enter EMPTY for the empty set.)
The set of natural numbers x that satisfy x + 2 = 1
Mathematics
1 answer:
castortr0y [4]3 years ago
4 0

Answer:

The empty set \{ \}

Step-by-step explanation:

Roster method is simply listing explicitly all the elements in the set, one by one (writing them between two curly brackets, and separating them through commas).

We want then to list explicitly all the elements in the following set:

The set of natural numbers x that satisfy x+2=1.

So, first we have to figure out which numbers are in that set. The set is made ONLY of those natural numbers x, that when you add 2 to them, you get 1. Clearly no natural number has that property (since the only number that would give us 1 when adding 2 to it, is the number -1, which is NOT a natural number). So there aren't any numbers at all in that set. So if we were to list them, we'd just list nothing inside the set:

\{ \} (which is just the empty set)

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Are short-term investments less prone to price validity than long-term investments true or false
gizmo_the_mogwai [7]

Answer:

yes its true

Step-by-step explanation

its true

5 0
2 years ago
Find all the solutions for the equation:
Contact [7]

2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0

Divide both sides by x^2\,\mathrm dx to get

2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}

Substitute v(x)=\dfrac{y(x)}x, so that \dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}. Then

x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}

x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}

The remaining ODE is separable. Separating the variables gives

\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x

Integrate both sides. On the left, split up the integrand into partial fractions.

\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}

\implies v^2+2v+1=a(v^2+1)+(bv+c)v

\implies v^2+2v+1=(a+b)v^2+cv+a

\implies a=1,b=0,c=2

Then

\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v

On the right, we have

\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C

Solving for v(x) explicitly is unlikely to succeed, so we leave the solution in implicit form,

\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C

and finally solve in terms of y(x) by replacing v(x)=\dfrac{y(x)}x:

\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C

\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}

7 0
3 years ago
make a box-and-whisker plot for the data. numbers of colors in a country's flag :3,2,2,4,4,3,6,3,5,3,4,1
Irina-Kira [14]

Answer:  see attachment

<u>Step-by-step explanation:</u>

Put the numbers in order from smallest to biggest:

1, 2, 2, 3, 3, 3,         3, 4, 4, 4, 5, 6

        ↓              ∨                ↓

       Q1        Median          Q3

Q1 median = (2+3)/2 = 2.5  <---- left side of box plot

Median = (3+3)/2 = 3          <---- Vertical Line of box plot

Q3 median = (4+4)/2 = 4    <---- right side of box plot

3 0
3 years ago
What is the value of x in the equation 5(4x-10)+10x=4(2x-3)+2(x-4) ?
MrRissso [65]
5(4x - 10) + 10x = 4(2x - 3) + 2(x - 4)
20x - 50 + 10x = 8x - 12 + 2x - 8
30x - 50 = 10x - 20
30x - 10x = -20 + 50
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x = 3/2 <==
6 0
3 years ago
Read 2 more answers
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Morgarella [4.7K]

Answer:

x < -4 or x > 8

On a number line, make two arrows:

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2) going towards right from 8 with an open circle at 8

Step-by-step explanation:

l -4 +2x l -3 > 9

l -4 +2x l > 12

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x < -4

On a number line, make two arrows:

1) going towards left from -4 with an open circle at -4

2) going towards right from 8 with an open circle at 8

8 0
3 years ago
Read 2 more answers
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