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damaskus [11]
3 years ago
6

Solve for the indicated variable 3x - y = 10; y

Mathematics
1 answer:
VashaNatasha [74]3 years ago
7 0

Answer:

y = 3x - 10

Step-by-step explanation:

Starting with 3x - y = 10, add y to both sides:

3x = 10 + y

Then subtract 10 from both sides:

3x - 10 = y

If you switch the sides of the equation, it can also be written as:

y = 3x - 10

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given y>0 and dy/dx=(3x^2+4x)/yif the point (1,√10) is on the graph relating x and y, then what is y when x=0
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\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2+4x}y\iff y\,\mathrm dy=(3x^2+4x)\,\mathrm dx
\implies\displaystyle\int y\,\mathrm dy=\int(3x^2+4x)\,\mathrm dx
\implies \dfrac12y^2=x^3+2x^2+C

When x=1 you have y=\sqrt{10}, so

\dfrac12(\sqrt{10})^2=1^3+2(1)^2+C\implies 5=1+2+C\implies C=2

and so the particular solution to the ODE is

\dfrac12y^2=x^3+2x^2+2

Then when x=0, you get

\dfrac12y^2=0^3+2(0)^2+2=2
\implies y^2=4
\implies y=2

where we omitted the negative root because it's given that y>0.
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