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lubasha [3.4K]
2 years ago
7

What is the sum of 1/2 and 1/3

Mathematics
2 answers:
Vesna [10]2 years ago
7 0
1/2+1/3=
3/6+2/6=5/6
The answer is 5/6.
stiks02 [169]2 years ago
4 0
The answer is 0.833333 repeating. you get that because
1/2=0.5  and 1/3=0.33 repeated then you add them together and get 0.8333 repeated.
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The hypotenuse of a 45°-45°-90° triangle measures 8 inches. What is the length of each leg?
LenaWriter [7]

Answer:

4.24, the hypotenuse is each leg multiplied by root 2, so you divide 6 by the root of 2

Step-by-step explanation:

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HELP!!! How do you do a diophantine equation? Plz help as soon as possible.
Annette [7]

Answer and Step-by-step explanation:

A Diophantine equation is simply an equation that relates whole numbers.

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4 0
3 years ago
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vovikov84 [41]
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3 0
3 years ago
Read 2 more answers
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
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