Hola necesito ayuda con esto

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago

(BRAINLIEST TO THE BEST ANSWER + PLEASE ANSWER PROPERLY, OR YOUR ANSWER WILL BE REPORTED. DON'T FORGET TO THOROUGHLY EXPLAIN.) J

KatRina [158]

Hey there!

In order to answer this question, it's easiest to start with one of the money amounts and work your way to the other ones. Keep in mind that, no matter what, the amount of collective money won't exceed 1.50, meaning you can't have two 1 dollar bills, etc. When I wrote it out, I did it in this order:

1 = 1 dollar
0.5 = Half dollar
0.25 = Quarter

1 + 0.5 = 1.50
1 + 0.25 + 0.25 = 1.50
0.5 + 0.5 + 0.5 = 1.50
0.5 + 0.5 + 0.25 + 0.25 = 1.50
0.5 + (0.25*4) = 1.50
0.25*6 = 1.50

Since no other combinations add up to 1.50, your answer is 6.

Hope this helped you out! :-)

4 0

3 years ago

What is the general form of the equation for the given circle centered at [0, 0)?

sergejj [24]

Answer:

x^2+y^2=r^2 is quation of circle whose centre is (0,0)

3 0

3 years ago

Whats the center and radius of (x+3)^2+(y-5)^2=64

Alexandra [31]

Answer:

The answer to your question is Center = (-3, 5) radius = 8

Step-by-step explanation:

Data

(x + 3)² + (y - 5)² = 64

Process

Because of the characteristics of the equation, we conclude that it is a circle.

Standard form of a circle:

(x - h)² + (y - k)² = r²

In this form of the equation, the center has coordinates (h, k), just change the signs.

And to get the radius, just get the square root of r²

h = -3

k = 5

r = 8

Center = (-3, 5)

radius = 8

5 0

3 years ago

The number 40 cannot be decomposed into prime factors True either False

dusya [7]

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{\textsf{\textbf{\underline{\underline{Answer : \:}}}}}}$

False, because the number 40 is a composite number, those that cannot be broken down are called Prime Numbers.

Prime numbers are numbers that are used to decompose any number, but prime numbers cannot be decomposed, since they only have two Divisors, which are 1 and itself.

$~\sf{Atte: Murdock:)}$

3 0

1 year ago

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Hola necesito ayuda con esto ​

Hola necesito ayuda con esto