We want to solve the following system of equations using the substitution method.

1 answer:

5 0

In the second equation you are given what Y equals, which is (-5x - 3). You would use this equation and plug it into the y value given in the first equation where it says 2y and solve

That would be

3x - 2(-5x - 3) = -6
3x + 10x + 6 = -6
3x + 10x = -6 - 6
13x = -12
X = -12/13

Then if you want to solve for Y you can use any equation and plug in the x-value found.

I’m going to use equation 2.

Y = -5x - 3
Y = -5(-12/13) - 3
Y = 4.615 - 3
Y = 1.615

(-12/13, 1.615)

Therefore the x-value is -12/13 and the y-value is 1.615.

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Bobby jo made 60 holiday cookies. 4/5 of the cookies are for her teachers and the rest are for her family .How many cookies are

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A sphere of radius r is cut by a plane h units above the equator, where

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Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$