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3 years ago

Read this excerpt from A Black Hole is NOT a Hole.

Mathematics

2 answers:

diamong [38]3 years ago

8 0

Answer:

I think the author is meaning to tell you to use your imagination to think about what will happen when you enter. You don’t know until you through it.

Step-by-step explanation:

Yuliya22 [10]3 years ago

3 0

Answer:

Step-by-step explanation:

Sorry if this is wrong its one of those edge quizes that dosnt tell you if you got it right or not.

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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

He simultaneous equations 5x + y = 21 x - 3y = 9 With working out

Elza [17]

Answer:

see explanation

Step-by-step explanation:

Given the 2 equations

5x + y = 21 → (1)

x - 3y = 9 → (2)

Multiply (1) by 3 and add the result to (2) to eliminate y term

15x + 3y = 63 → (3)

Add (2) and (3) term by term

(x + 15x) + )- 3y + 3y) = (9 + 63)

16x = 72 ( divide both sides by 16 )

x = 4.5

Substitute x = 4.5 into (1) for corresponding value of y

22.5 + y = 21 ( subtract 22.5 from both sides )

y = - 1.5

Solution is (4.5, - 1.5 )

8 0

3 years ago

The fraction 7/9 produces a repeating decimal 0.7 A. true B. false

Elis [28]

Answer:

false for 0.7 it would have to be 7/10

Step-by-step explanation:

7 0

3 years ago

Simplify the expression.

snow_tiger [21]

The answer to the question is d.

8 0

3 years ago

Suppose a city with population 800,000 has been growing at a rate of 5% per year. If this rate continues, find the population of

AleksandrR [38]

Answer:

2,340,208.5 or 2,340,209

Step-by-step explanation:

You could answer this by multiplying the current population of 800,000 by 1.05 (1.05 represents the annual growth rate of population) and do that 22 times. But that would take a while.

we got 1.05 because the formula says that y = a( 1 + r ) power t

for a we will give it 1

so y = 1 ( 1 + 0.05) power t

it will give us 1.05

to do this faster, you would first calculate (1.05)22and then multiply this by 800,000.

So, 800,000 x (1.05)22 = 2,340,208.5 or 2,340,209

5 0

2 years ago