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3 years ago

What it bigger 40% of 90 or half of 80

1 answer:

Ber [7]3 years ago

6 0

$p\%=\frac{p}{100}\\\\40\%=\frac{40}{100}=\frac{4}{10}=0.4\\\\40\%\ of\ 90=0.4\cdot90=\fbox{36}\\\\half=\frac{1}{2}=0.5\\\\half\ of\ 80=0.5\cdot80=\fbox{40}\\\\therefore,\ bigger\ is\ \boxed{half\ of\ 80}$

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sashaice [31]

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

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

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

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3 years ago

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Step-by-step explanation:

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densk [106]

Answer: OPTION B

Step-by-step explanation:

The equation of a line that passes through the origin is the following:

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3 years ago

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