What is the y-intercept of the function described in the table?

Organizers are launching fireworks over the top of the Washington

barxatty [35]

<h3>sorry I don't know </h3>

<h2>sorry I don't know </h2>

7 0

3 years ago

Surface integrals using a parametric description. evaluate the surface integral \int \int_{s} f(x,y,z)dS using a parametric desc

DiKsa [7]

You can parameterize $S$ using spherical coordinates by

$\vec s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial\vec v}\times\dfrac{\partial\vec s}{\partial\vec u}=36\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle$

(I use $\vec s_v\times\vec s_u$ to avoid negative signs. The orientation of the normal vector doesn't matter for a scalar surface integral; you could just as easily use $\vec s_u\times\vec s_v=-(\vec s_v\times\vec s_u)$ .)

Then

$f(x,y,z)=f(6\cos u\sin v,6\sin u\sin v,6\cos v)=36\sin^2v$

and the integral of $f$ over $S$ is

$\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\int_0^{\pi/2}\int_0^{2\pi}36\sin^2v\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}(36\sin^2v)(36\sin v)\,\mathrm du\,\mathrm dv$

$=\displaystyle2592\pi\int_0^{\pi/2}\sin^3v\,\mathrm dv=\boxed{1728\pi}$

7 0

3 years ago

Determine whether each of the binary relation R defined on the given sets A is reflexive, symmetric, antisymmetric, or transitiv

ki77a [65]

Answer:

In explanation

Please let me know if something doesn't make sense.

Step-by-step explanation:

a)

*This relation is not reflexive.

0 is an integer and (0,0) is not in the relation because 0(0)>0 is not true.

*This relation is symmetric because if a(b)>0 then b(a)>0 since multiplication is commutative.

*This relation is transitive.

Assume a(b)>0 and b(c)>0.

Note: This means not a,b, or c can be zero.

Therefore we have abbc>0.

Since b^2 is positive then ac is positive.

Since a(c)>0, then (a,c) is in R provided (a,b) and (b,c) is in R.

*The relation is not antisymmretric.

(3,2) and (2,3) are in R but 3 doesn't equal 2.

b)

*This relation is reflective.

Since a^2=a^2 for any a, then (a,a) is in R.

*The relation is symmetric.

If a^2=b^2, then b^2=a^2.

*The relation is transitive.

If a^2=b^2 and b^2=c^2, then a^2=c^2.

*The relation is not antisymmretric.

(1,-1) and (-1,1) is in the relation but-1 doesn't equal 1.

c)

*The relation is reflexive.

a/a=1 for any a in the naturals.

*The relation is not symmetric.

Wile 4/2 is an integer, 2/4 is not.

*The relation is transitive.

If a/b=z and b/c=y where z and y are integers, then a=bz and b=cy.

This means a=cyz. This implies a/c=yz.

Since the product of integers is an integer, then (a,c) is in the relation provided (a,b) and (b,c) are in the relation.

*The relation is antisymmretric.

Assume (a,b) is an R. (Note: a,b are natural numbers.) This means a/b is an integer. This also means a is either greater than or equal to b. If b is less than a, then (b,a) is not in R. If a=b, then (b,a) is in R. (Note: b/a=1 since b=a)

6 0

3 years ago

What is the area of the figure:

Luba_88 [7]

For this case we have that the area of the triangle is given by:

$A = \frac {b * h} {2}$