END OF UNIT MATH TEST PLEASE HELP

Mathematics

1 answer:

maksim [4K]2 years ago

8 0

Answer:

b and e

Step-by-step explanation:

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Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 =

sashaice [31]

Let $R$ be the solid. Then the volume is

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

which follows from the facts that

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$
(by computing the Jacobian)

and

$z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8$
(we take the positive solution, since it's clear that $R$ lies above the $x$ - $y$ plane)
$r^2+z^2=72\implies z=\pm\sqrt{72-r^2}$
(again, taking the positive root for the same reason)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr$
$=\dfrac{32(27\sqrt2-35)\pi}3$

7 0

3 years ago

What are the vertical and horizontal asymptote of f(x)=2x/x-1

makvit [3.9K]

F(x)=2x/(x-1)

Denominator cannot be equal 0.
x-1=0
x=1 is vertical asymptote.

To find horizontal asymptote
we need divide the coefficients before x
y=2x/x=2,because when x becomes really big (x---->∞) , value of 1 will not influence the result .
y=2 is horizontal asymptote

6 0

3 years ago

Nvm bro i got the answer

KIM [24]

Ok and have a nice day

6 0

2 years ago

Which are the soultions of the quadratic equation? x^2=7x+4

Vesnalui [34]

Answer:

Answer in image below

Step-by-step explanation: