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OLga [1]
3 years ago
10

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.

6 0
3 years ago
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Perimeter is 22 units , one side is 6 units
kvv77 [185]
Is it asking for the perimeter of a square?
4 0
3 years ago
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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}

7 0
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}}

\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=e^u\sqrt{u^2+v^2+e^{2u}}

So we have

\mathrm dS=\boxed{e^u\sqrt{u^2+v^2+e^{2u}}\,\mathrm du\,\mathrm dv}

8 0
3 years ago
Solve the following inequality algebraically. <br><br> | x-8 | &gt; 4
bulgar [2K]

Answer:

x>12

This is the answer

6 0
3 years ago
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