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2 years ago

Nevermind i dont need help

Mathematics

1 answer:

joja [24]2 years ago

3 0

Perf!!! Thanks for the points

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What is an apparent solution

NNADVOKAT [17]

Answer:

an apparent solution that must be rejected because it does not satisfy the original equation.

Step-by-step explanation:

7 0

3 years ago

PLS SOMEONE HELP I DO NOT UNDERSTAND PLS AND TY ASAP PLS PLS I WILL GIVE BRAINLIST AND PLS DONT BE A BOT

Kruka [31]

Answer:

35°

Step-by-step explanation:

The central arc is equal to the arc that subtends it, then

arc AC = 75° and

BC = AC - AB = 75° - 40° = 35°

8 0

2 years ago

Answer my question #4

Vinil7 [7]

Answer:

11. 432,500 12. KF & ND

Step-by-step explanation:

11.

80,000 + 75,000 + 105,000 + 82,500 + 90, 000 = 432,500 people

( it says "about" so it doesn't have to be exact but a reasonable guess)

12. The two stadiums most similar in attendance is Kyle Field and Notre Dame

5 0

2 years ago

Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x

kykrilka [37]

Parameterize $S$ by

$\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take a normal vector to $S$ ,

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k$

which has norm

$\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v$

Then the integral of $f(x,y,z)=x^2+y^2$ over $S$ is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv$

$=\displaystyle36^2\int_0^{\pi/2}\int_0^{2\pi}\sin^3v\,\mathrm du\,\mathrm dv=\boxed{1728\pi}$

6 0

3 years ago

What is the square root of 64y16? 4y4 4y8 8y4 8y8

aliya0001 [1]

It is 8y4 becuze 8*2=64 and 4*2=16 and u *2 becuze *2 means square root sorry if I'm wrong :)

7 0

3 years ago

Read 2 more answers