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valkas [14]
3 years ago
5

Integrate ​G(x,y,z)equalsz over the parabolic cylinder yequalszsquared​, 0less than or equalsxless than or equals2​, 0less than

or equalszless than or equalsStartFraction StartRoot 15 EndRoot Over 2 EndFraction . Write the integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S Upper G (x comma y comma z )d sigma as a double integral. ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S Upper G (x comma y comma z )d sigmaequalsIntegral from nothing to nothing Integral from nothing to nothing nothing dz dx.
Mathematics
1 answer:
yKpoI14uk [10]3 years ago
5 0

I gather you're supposed to compute the integral of G(x,y,z)=z over a surface S that is the part of the parabolic cylinder y=z^2 with 0\le x\le2 and 0\le z\le\frac{\sqrt{15}}2.

We can parameterize S by

\vec s(x,z)=x\,\vec\imath+z^2\,\vec\jmath+z\,\vec k

with the given constraints on x and z. Take the normal vector to S to be

\vec s_x\times\vec s_z=-\vec\jmath+2z\,\vec k

so that the surface element is

\mathrm dS=\|\vec s_x\times\vec s_z\|\,\mathrm dx\,\mathrm dz=\sqrt{1+4z^2}\,\mathrm dx\,\mathrm dz

Then in the integral, we have

\displaystyle\iint_Sz\,\mathrm dS=\int_0^2\int_0^{\sqrt{15}/2}z\sqrt{1+4z^2}\,\mathrm dz\,\mathrm dx=\boxed{\frac{21}2}

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Y=4 - 5x<br> a. m=4, b=-5<br> b. m=-5, b=4<br><br> help
vredina [299]

Answer:

b

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + b ( m is the slope and b the y- intercept )

y = 4 - 5x or y = - 5x + 4 ← is in slope- intercept form

with m = - 5 and b = 4

3 0
2 years ago
I need to know what is the slope line passing thorough the points(0,-5) and (4,2
julsineya [31]
Ok so slope formula is y2 - y1/x2 - x1 so it would be 2- -5 which is -7 and 4 - 0 which is 4 so your answer would be -7/4. hope this helps i would appreciate a brainliest
6 0
2 years ago
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If the sumof two numbers,n and m,is rational,which statement is true?
Nataly_w [17]

Option B:

Both n and m must be rational.

Solution:

Given information:

Sum of two numbers, n and m are rational.

<u>To find which statements are true:</u>

Option A: Both n and m may be rational but do not have to be.

It is not true because n and m given is rational.

It have to be rational.

Option B: Both n and m must be rational.

Yes, n and m must be rational then only the sum of numbers are rational.

It is true.

Option C: Both n and m must be irrational.

Sum of irrationals will be sometimes irrational and sometimes can't add.

So it is not true.

Option D: One number is rational and the other is irrational.

Rational and irrational cannot be add.

So it is not true.

Option B is true.

Both n and m must be rational.

5 0
3 years ago
The reduced fraction of 1902/21
mars1129 [50]
So i got 90 4/7 reduced in fractions
5 0
3 years ago
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A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per
Tems11 [23]

Answer:

The pressure is changing at \frac{dP}{dt}=3.68

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} and we want to find at what rate is the pressure changing.

The equation that model this situation is

PV^{1.4}=k

Differentiate both sides with respect to time t.

\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)

Apply this rule to our expression we get

V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0

Solve for \frac{dP}{dt}

V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}

when P = 23 kg/cm2, V = 35 cm3, and \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} this becomes

\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68

The pressure is changing at \frac{dP}{dt}=3.68.

7 0
3 years ago
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