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2 years ago

A rectangular tank measuring 35 cm by 28 cm by 16 cm is 2/5 filled with

Mathematics

1 answer:

Alina [70]2 years ago

7 0

Answer:

Amount of water left in tank = 440 cm³

Step-by-step explanation:

Given:

Sides of tank respectively = 35 cm , 28 cm , 16 cm

Water level in tank = 2/5

Side of a cubic tank = 18 cm

Find:

Amount of water left in tank

Computation:

Volume level in rectangular tank = [2/5][35 x 28 x 16]

Volume level in rectangular tank = [2/5][15,680]

Volume level in rectangular tank = [31,360/5]

Volume level in rectangular tank = 6,272 cm³

Volume of cube = side³

Volume of cube = 18³

Volume of cube = 5,832 cm³

Amount of water left in tank = Volume level in rectangular tank - Volume of cube

Amount of water left in tank = 6,272 - 5,832

Amount of water left in tank = 440 cm³

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2 years ago

If 2.5 mol of dust particles were laid end to end along the equator, how many times would they encircle the planet? The circumfe

Natalka [10]

Answer:

They encircle the planet $3.76\times 10^{11}$ times.

Step-by-step explanation:

Consider the provided information.

We have 2.5 mole of dust particles and the Avogadro's number is $6.022\times 10^{23}$

Thus, the number of dust particles is:

$2.5\times 6.022\times 10^{23}=15.055\times 10^{23}$

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Convert the measurement in meters.

Diameter: $10\mu m\times \frac{10^{-6}m}{\mu m} =10^{-5}m$

If we line up the particles the distance they could cover is:

$15.055\times 10^{23}\times 10^{-5}=15.055\times 10^{18}=1.5055\times 10^{19}$

Circumference in meters:

$40,076km\times \frac{1000m}{1km}=40,076,000 m$

Therefore,

$\frac{1.5055\times 10^{19}}{40,076,000} = 3.76\times 10^{11}$

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2 years ago

What additional information could you use to show that ΔSTU ≅ ΔVTU using SAS? Check all that apply.

Ugo [173]

I believe the correct answers are:

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2 years ago

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sattari [20]

2 = $1.45
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2 years ago

ratelena [41]

Answer:

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Step-by-step explanation:

Given : There are 10 people in the party .

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Solution:

We are given that there are 10 people in the party

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To find the no. of possible handshakes between 10 people we will use combination over here

Formula : $^nC_r=\frac{n!}{r!(n-r)!}$