Question 2 (1 point) Solve 2d - 7= 8-34. O a d- 5 Oь d = -15 Ос d=-1 5 Od Submit<br>

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

Which shows a graph of a linear equation in standard form Ax + By = C, where A = 0, B is positive, and C is negative? A coordina

IrinaK [193]

Answer:

A coordinate plane with a horizontal line passing through (negative 4, negative 3), (0, negative 3) and (4, negative 3).

Step-by-step explanation:

Ax + By = C, where A = 0, B is positive, and C is negative

By= C or y= C/B which is negative

Since A is zero and C is negative, the line is parallel to x- axis and has negative value for y, so it is a horizontal line below zero

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A coordinate plane with a vertical line passing through (1, negative 4), (1, 0) and (1, 4).

no, the line is horizontal

A coordinate plane with a vertical line passing through (negative 2, negative 4), (negative 2, 0) and (negative 2, 4).

no, the line is horizontal

A coordinate plane with a horizontal line passing through (negative 4, 4), (0, 4) and (negative 4, 4).

no, the y-values should be negative

A coordinate plane with a horizontal line passing through (negative 4, negative 3), (0, negative 3) and (4, negative 3).

yes, this is horizontal line below zero

8 0

3 years ago

Read 2 more answers

Perez’s Ski Shop made a profit of $32,000 last year. This year, the profits are 250 percent of last year’s profits. How much of

Tanzania [10]

Answer:

$80,000 I think is right

3 0

3 years ago

Read 2 more answers

Slove system of equation.<br><br> 2x - 6y = 8<br> -3x + 9y = 12

Bond [772]

Answers:

2x - 6y = 8 -> x = 4 + 3y

-3x + 9y = 12 -> x = -4 + 3y

7 0

2 years ago

A pitcher holds 3/4 gallons of kool-aid. Martin wants to pour equal amounts of kool- aid into 6 smaller cotainers. How much kool

statuscvo [17]

Answer:

$\dfrac{1}{8}\ gallons$

Step-by-step explanation:

Amount of kool - aid that the pitcher can hold = $\frac{3}4$ gallons

There are 6 smaller containers in which this kool - aid is to be adjusted.

Also, the volume of the smaller containers is equal to each other i.e. the containers are identical to each other in terms of volume.

To find:

The amount of kool - aid that each container can hold?

Solution:

Volume of all the smaller containers combined = Volume of the kool - aid available

Let the volume of one smaller container = $V$

$V + V + V + V + V + V = \dfrac{3}{4}\ gallons$

$\Rightarrow 6V = \dfrac{3}{4}\ gallons\\\Rightarrow V = \dfrac{3}{4\times 6 }\ gallons\\\Rightarrow V=\bold{\dfrac{1}{8}\ gallons}$

$\frac{1}{8}\ gallons$ of kool - aid is the amount that each smaller can hold.

8 0

3 years ago

Question 2 (1 point) Solve 2d - 7= 8-34. O a d- 5 Oь d = -15 Ос d=-1 5 Od Submit​

Question 2 (1 point) Solve 2d - 7= 8-34. O a d- 5 Oь d = -15 Ос d=-1 5 Od Submit