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3 years ago

Solve the formula V=pir^2h for r PLEAASSSEEE HELP

Mathematics

2 answers:

otez555 [7]3 years ago

7 0

Answer:

Step-by-step explanation:

So we have the formula:

$V=\pi r^2h$

And we want to solve it for r.

So, let's first divide both sides by π and h. This will cancel out the right side:

$r^2=\frac{V}{\pi h}$

Now, take the square root of both sides:

$r=\sqrt{\frac{V}{\pi h}}$

And we're done!

Our answer is B.

I hope this helps!

Tasya [4]3 years ago

6 0

Answer:

B. $r=\sqrt{\frac{v}{\pi h}$

Step-by-step explanation:

We are given the formula:

$V=\pi r^2 h$

and asked to solve for $r$ . Therefore, we must isolate $r$ on one side of the equation.

$\pi$ and $h$ are both being multiplied by r². The inverse of multiplication is division. Divide both sides of the equation by $\pi h$ .

$\frac{v}{\pi h} =\frac{\pi r^2 h}{\pi h}$

$\frac{v}{\pi h} =r^2$

r is being squared. The inverse of a square is a square root. Take the square root of both sides of the equation.

$\sqrt{\frac{v}{\pi h}} =\sqrt{r^2}$

$\sqrt{\frac{v}{\pi h}}=r$

$r=\sqrt{\frac{v}{\pi h}$

Therefore, the correct answer is B. $r=\sqrt{\frac{v}{\pi h}$

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Determine the values of the constants r and s such that i(x, y) = x rys is an integrating factor for the given differential equa

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$\underbrace{y(7xy^2+6)}_{M(x,y)}\,\mathrm dx+\underbrace{x(xy^2-1)}_{N(x,y)}\,\mathrm dy=0$

For the ODE to be exact, we require that $M_y=N_x$ , which we'll verify is not the case here.

$M_y=21xy^2+6$
$N_x=2xy^2-1$

So we distribute an integrating factor $i(x,y)$ across both sides of the ODE to get

$iM\,\mathrm dx+iN\,\mathrm dy=0$

Now for the ODE to be exact, we require $(iM)_y=(iN)_x$ , which in turn means

$i_yM+iM_y=i_xN+iN_x\implies i(M_y-N_x)=i_xN-i_yM$

Suppose $i(x,y)=x^ry^s$ . Then substituting everything into the PDE above, we have

$x^ry^s(19xy^2+7)=rx^{r-1}y^s(x^2y^2-x)-sx^ry^{s-1}(7xy^3+6y)$
$19x^{r+1}y^{s+2}+7x^ry^s=rx^{r+1}y^{s+2}-rx^ry^s-7sx^{r+1}y^{s+2}-6sx^ry^s$
$19x^{r+1}y^{s+2}+7x^ry^s=(r-7s)x^{r+1}y^{s+2}-(r+6s)x^ry^s$
$\implies\begin{cases}r-7s=19\\r+6s=-7\end{cases}\implies r=5,s=-2$

so that our integrating factor is $i(x,y)=x^5y^{-2}$ . Our ODE is now

$(7x^6y+6x^5y^{-1})\,\mathrm dx+(x^7-x^6y^{-2})\,\mathrm dy=0$

Renaming $M(x,y)$ and $N(x,y)$ to our current coefficients, we end up with partial derivatives

$M_y=7x^6-6x^5y^{-2}$
$N_x=7x^6-6x^5y^{-2}$

as desired, so our new ODE is indeed exact.

Next, we're looking for a solution of the form $\Psi(x,y)=C$ . By the chain rule, we have

$\Psi_x=7x^6y+6x^5y^{-1}\implies\Psi=x^7y+x^6y^{-1}+f(y)$

Differentiating with respect to $y$ yields

$\Psi_y=x^7-x^6y^{-2}=x^7-x^6y^{-2}+\dfrac{\mathrm df}{\mathrm dy}$
$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

Thus the solution to the ODE is

$\Psi(x,y)=x^7y+x^6y^{-1}=C$