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vladimir1956 [14]
3 years ago
13

A 5 gram ball has a volume of 2.5 milliliters. What is the density of the ball?

Mathematics
1 answer:
lisabon 2012 [21]3 years ago
7 0

Answer:

2gramm/millimeters

Step-by-step explanation:

From the question we were given 5

gram ball with a volume of 2.5 milliliters.

Then What is the density of the ball?

Density of a substance is the mass of the substance per it's unit volume

It can be expressed below as

Density= mass/ volume

Mass= 5gram

Volume= 2.5millimetre

Then substitute we have

Density=5/2.5

=2g/mm

Therefore, density of the ball is 2gramm/millimeters

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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} }

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Point m is the midpoint of kL. M(1,-1) and L(8,-7. What are the coordinates of point k?
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Answer: (-6, 5)

<u>Step-by-step explanation:</u>

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Separate the x's and y's and solve them individually:

\dfrac{x_k+x_L}{2}=x_m\qquad \qquad \qquad \qquad\dfrac{y_k+y_L}{2}=y_m\\\\\\\dfrac{x_k+8}{2}=1\qquad \qquad \qquad \qquad \qquad \dfrac{y_k-7}{2}=-1\\\\\\x_k+8=2\qquad \qquad \qquad \qquad \qquad y_k-7=-2\\\\\\x_k\quad =-6 \qquad \qquad \qquad \qquad \qquad y_k\qquad =5

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