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marin [14]
4 years ago
9

What is the volume of the sphere in terms of ? V = 8 in. 3

Mathematics
1 answer:
nikdorinn [45]4 years ago
7 0
Couldn’t find a way to type it but hope it helps!!

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How to write in inqvality
Kitty [74]

Answer:

If you mean "How do I write an inequality?", the answer is COEFFICIENTxVARIABLE {any symbol} COEFFICIENTxVARIABLE >/< INTEGER

In other words, you should have something like this: -3x + 4v < 63

Or this: 51x - 16v > 11

Or anything along those lines.

8 0
3 years ago
Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
Harrizon [31]

Answer:

The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

f_z = -ye^{yz}

we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

3 0
4 years ago
Please help me don't understand :(
qaws [65]

Answer:

I think it's the last one. Sorry if I'm wrong

8 0
3 years ago
HELP PLEASE! HELP! HELP!
kifflom [539]
I think it is D 1.8. hope this help's
5 0
3 years ago
How do you do the equation 2 to the second power over 2 to the 3rd power
padilas [110]

2 to the second power is 2 x 2 (4) , and 2 to the 3rd power is 2 x 2 x 2 (8)

3 0
3 years ago
Read 2 more answers
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