3 years ago

Without solving what can you tell about the solution

Mathematics

1 answer:

nlexa [21]3 years ago

8 0

Answer:

n has to be negative

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Suppose that the time to complete a test is bell-shaped distributed with mean 45 minutes and standard deviation 15 minutes. if t

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Use a surface integral to ﬁnd the surface area of the portion of the sphere xUse a surface integral to ﬁnd the surface area of t

brilliants [131]

Parameterize this surface (call it $S$ ) by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\dfrac\pi4$ . The limits on $u$ should be obvious. We find the upper limit for $v$ by solving for $v$ where the sphere and cone intersect:

$\begin{cases}x^2+y^2+z^2=1\\z=\sqrt{x^2+y^2}\end{cases}\implies x^2+y^2=\dfrac12$

$\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\dfrac12$

$\implies\sin^2v=\dfrac12$

$\implies\sin v=\dfrac1{\sqrt2}\implies v=\dfrac\pi4$

Take the normal vector to $S$ to be

$\vec r_u\times\vec r_v=-\cos u\sin^2v\,\vec\imath-\sin u\sin^2v\,\vec\jmath-\cos v\sin v\,\vec k$

(orientation does not matter here)

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{\pi/4}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv$

$=\displaystyle2\pi\int_0^{\pi/4}\sin v\,\mathrm dv=\boxed{(2-\sqrt2)\pi}$

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3 years ago

The oblique pyramid has a square base with an edge length of 5 cm. The height of the pyramid is 7 cm.

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Answer:

58 cm³

Step-by-step explanation:

the volume (V) of a pyramid is calculated using the formula

V = $\frac{1}{3}$ × area of base × height

= $\frac{1}{3}$ × 5² × 7 ≈ 58 cm³

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Answer: it’s compressed horizontally

Step-by-step explanation:

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The leg of a right triangle are 3 units and 7 units. What is the length of the hypotenuse? Round your answer to the nearest hund

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Step-by-step explanation:

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