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3 years ago

Solve the system of linear equations.

Mathematics

1 answer:

icang [17]3 years ago

7 0

Answer: The set does not have a solution

Step-by-step explanation:

Adding Equations 1 & 3 we get 5x = 7. This gives x = 7/5

Putting this value of x in eq. 2 we get

-2y + 2z = -1-(7/5) or

2y - 2z = 12/5 or 5y - 5z = 6

Multiplying eq. 1 by 2 we get

4x + 2y - 2z = 6

adding this with eq. 2 we get 5x = 5 or x = 1

As the common solution for x from equations 1&3 does not satisfy eq. 1&2 it comes out that the three equations do not have a common solution.

Same can be verified by using different sets of two equations also.

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3 years ago

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2 years ago

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Answer:

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2 years ago

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Anika [276]

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

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

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

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3 years ago

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Lena [83]

Answer:

$55.24

Step-by-step explanation:

Add $32.78 + $27.24 + $5.22

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