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3 years ago

WILL GIVE BRAINLIEST

Mathematics

2 answers:

Gelneren [198K]3 years ago

5 0

Answer:

the answer it's B we got the same test

PUT THE B

irina1246 [14]3 years ago

5 0

Answer:

mm option a is right

Step-by-step explanation:

bcz there is only 1 zero less in weight of elephant as compared to whale hope it may help u

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How much is an acute traingle?

tangare [24]

Answer:

Step-by-step explanation:

An acute triangle (or acute-angled triangle) is a triangle with three acute angles (less than 90°). An obtuse triangle (or obtuse-angled triangle) is a triangle with one obtuse angle (greater than 90°) and two acute angles.

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Integrate the following

enot [183]

I suppose you mean to have the entire numerator under the square root?

$\displaystyle\int_2^4\frac{\sqrt{x^2-4}}{x^2}\,\mathrm dx$

We can use a trigonometric substitution to start:

$x=2\sec t\implies\mathrm dx=2\sec t\tan t\,\mathrm dt$

Then for $x=2$ , $t=\sec^{-1}1=0$ ; for $x=4$ , $t=\sec^{-1}2=\frac\pi3$ . So the integral is equivalent to

$\displaystyle\int_0^{\pi/3}\frac{\sqrt{(2\sec t)^2-4}}{(2\sec t)^2}(2\sec t\tan t)\,\mathrm dt=\int_0^{\pi/3}\frac{\tan^2t}{\sec t}\,\mathrm dt$

We can write

$\dfrac{\tan^2t}{\sec t}=\dfrac{\frac{\sin^2t}{\cos^2t}}{\frac1{\cos t}}=\dfrac{\sin^2t}{\cos t}=\dfrac{1-\cos^2t}{\cos t}=\sec t-\cos t$

so the integral becomes

$\displaystyle\int_0^{\pi/3}(\sec t-\cos t)\,\mathrm dt=\boxed{\ln(2+\sqrt3)-\frac{\sqrt3}2}$

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3 years ago

Compute the differential of surface area for the surface S described by the given parametrization.

AysviL [449]

With $S$ parameterized by

$\vec r(u,v)=\langle e^u\cos v,e^u\sin v,uv\rangle$

the surface element $\mathrm dS$ is

$\mathrm dS=\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

We have

$\dfrac{\partial\vec r}{\partial u}=\langle e^u\cos v,e^u\sin v,v\rangle$

$\dfrac{\partial\vec r}{\partial v}=\langle -e^u\sin v,e^u\cos v,u\rangle$

with cross product

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle ue^u\sin v-ve^u\cos v,-ve^u\sin v-ue^u\cos v,e^{2u}\cos^2v+e^{2u}\sin^2v\rangle$

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\langle e^u(u\sin v-v\cos v),-e^u(v\sin v+u\cos v),e^{2u}\rangle$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{e^{2u}(u\sin v-v\cos v)^2+e^{2u}(v\sin v+u\cos v)^2+e^{4u}}$