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2 years ago

Make 10 equations for 50,000

Mathematics

2 answers:

Kruka [31]2 years ago

4 0

Answer:

1. 50 * 1,000 = 50,000

2. 49,001 + 999 = 50,000

3. 5 * 10,000 = 50,000

4. 100,000/2 = 50,000

5. 20,000 + 30,000 = 50,000

6. 50,000/1 = 50,000

7. 90,000 - 40,000 = 50,000

8. 10 + 49,990 = 50,000

9. 5,000 * 10 = 50,000

10. 1,000,000/20 = 50,000

o-na [289]2 years ago

3 0

Answer/Step-by-step explanation:

These are simple math equations, so I do not think that there is any explanations needed.

1) 10 x 5,000

2) 5 x 10,000

3) 500 x 100

4) 250 x 200

5) 625 x 80

6) 2,500 x 20

7) 50 x 1,000

8) 25 x 2,000

9) 60,000 - 10,000

10) 100,000 / 2

I hope this helped!!

~Penny

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Sliva [168]

Answer:

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2 years ago

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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)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

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