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2 years ago

13

what is the mass of an object that has a density of 0.7g/cm^3 and a volume of 8cm^3? (The density of an object has the equation

d= m/v)

1 answer:

UNO [17]2 years ago

4 0

Mass=5.6
Find this by plugging your values into the density formula and solving for M.

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Solve this two step equation = 6x + 8 = 74. Please solve for x.

AysviL [449]

6x + 8 = 74

-8 -8

6x = 66

/6 /6

x = 11

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3 years ago

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Your bill at a café is $26.50. You want to leave a 20% tip for your good service. What is the amount of the tip and the total of

Licemer1 [7]

Your tip total will be $5.30 and your bill total will be $31.80 after tip.

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Match the given angle pair to their relationship using the figure beloe

hoa [83]

Answer:

Vertical: 1

corresponding: 3

alternate interior: 2

alternate exterior: 4

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2 years ago

A farmer from Whidbey Island: A man claims his whalebone can detect water. You design an experiment using a set of 4 cans, 3 emp

tatyana61 [14]

Answer:

The value is $E(x) = 1.25$

Step-by-step explanation:

From the question we are told that

The number of cans is n = 4

The number of can that are empty is N = 3

The number of can filled with water is k = 1

The number number of sets of cans is w = 5

Generally probability of detecting the correct can is mathematically represented as

$p = \frac{k}{n}$

=> $p = \frac{1}{4}$

=> $p = 0.25$

Generally probability of not detecting the correct can is mathematically represented as

$q = 1- p$

$q = 1- 0.25$

$q = 0.75$

Generally the number of cans expected of the farmer to correctly identify by chance is mathematically represented as

$E(x) = w * p$

=> $E(x) = 5 * 0.75$

=> $E(x) = 1.25$

5 0

2 years ago

A vertical right circular cylindrical tank has height h=8 feet high and diameter d=6 feet. It is full of kerosene weighing 50 po

Mariana [72]

Answer:

The work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

Step-by-step explanation:

The work is defined by:

$W=\int dFdx$ (1)

The force here will be the product between the volume and the kerosene weighing, so we have :

$dF=\pi R^{2}dy*50$

This force will be in-lbs.

Where R is the radius (3 feet)

Then using (1), we have:

$W=\int \pi R^{2}dy*50(8-y)$

Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.

$W=\int_{0}^{8} \pi 3^{2}dy*50(8-y)$

$W=450\pi \int_{0}^{8}dy(8-y)$

$W=450\pi(\int_{0}^{8} 8dy-\int_{0}^{8} ydy)$

$W=450\pi(8y|_{0}^{8} -\frac{y^{2}}{2}|_{0}^{8})$

$W=450\pi(8*8 -\frac{8^{2}}{2})$

$W=450\pi(64 -\frac{64}{2})$

Therefore, the work done to pump all of the kerosene from the tank to an outlet is $W=45238.9\: J$

I hope it helps you!

6 0

2 years ago