To add/sub decimals you must...

Mathematics

1 answer:

kati45 [8]3 years ago

8 0

Answer:

Line them up

Step-by-step explanation:

it is much more easier to line up decimals to add and subtract

Plzzz give me brainlest and say thanks

You might be interested in

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

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