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sveticcg [70]
3 years ago
14

A number line going from 0 to 3 in increments of 3.

Mathematics
2 answers:
Alexandra [31]3 years ago
6 0

Answer:

4

Step-by-step explanation:

did it on edge

butalik [34]3 years ago
3 0

Answer:

2 divided by 1/2 is 4.

Step-by-step explanation:

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Is 3/8 estimated to be closer to 0 or 1?
kolezko [41]
The cut-off point is 0.5: higher than this would be closer to 1, and lower would be closer to 0

and 3/8 is lower than 0.5: we know this because a bigger number, 4/8 is equal to 0.5

so it's closer to 0!
3 0
3 years ago
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A first number plus twice a second number is 9. Twice the first number plus the second totals 27. Find the numbers
SOVA2 [1]
Create a a system of 2 equations: 2x + y = 27 and 2y + x = 9, with x being the first number and y being the second. solve for the two variables and you'll get the first number is 15 and the second is -3.
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3 years ago
Blank divided by blank equals 6 what is the two blanks
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2 years ago
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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
A cylinder is sliced by a plane that is not parallel to the base and intersects only the lateral surface. the resulting cross se
Oxana [17]

The cylinder that is sliced by the plane which is not parallel to the base of the cylinder and intersects only at the lateral surface is called an ellipse.

<h3>What is the formation of an Ellipse in a Cylinder?</h3>

When a cylinder with a longitudinal axis (z) is sliced at an angle to the xy plane, an elliptical shape is created.

In a circle, every plane crosses a circular cylinder perpendicularly to its axis. A plane that is neither at right angles to the axis nor parallel to it, crosses the cylinder in an elliptical shape.

Learn more about an ellipse here:

brainly.com/question/26679189

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