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lisabon 2012 [21]
2 years ago
6

How many cubic blocks of side length

Mathematics
1 answer:
Andrei [34K]2 years ago
8 0

Answer:

To find how many blocks it would take in each of these scenarios, you could find the total volume and then divide by the volume of each smaller cube.

1.  3/7 x 3/7 x 3/7 = 27/343 cubic inches

    1/7 x 1/7 x 1/7 = 1/343 cubic inches

27/343 cubic inches  divided by 1/343 cubic inches = 27 cubes.

2.  3/7 x 1/7 x 3/7 = 9/343 cubic inches

9/343 cubic inches divided into groups of 1/343 cubic inches is 9 cubes.

3.  2/6 x 2/6 x 2/6 = 8/216 cubic inches

8/216 cubic inches divided by 1/216 cubic inches is 8 cubes.

Another way to think of it would be for #1, each 3/7 of an inch would be 3 groups of 1/7, so 3 x 3 x 3 for each dimension would result in 27 cubes.

For #2, it would be 3 groups of 1/7, one group of 1/7 and another 3 groups of 1/7, so 3 x 3 is 9 cubes.

For #3 it would be 2 groups of 1/6 three times, so 2 x 2 x 2 = 8 cubes.

 

 

Step-by-step explanation:

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Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

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What is...<br><br> 7 4/7 + 14 4/5<br> Thank you so much!
SCORPION-xisa [38]
7+4/7+14+4/5
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7 0
3 years ago
Please help i put a picture below
PSYCHO15rus [73]

Answer:

Step-by-step explanation:

This is not nearly as threatening and scary as I first thought it was.  You must be in the section in Geometry where you are taught that perimeter of similar figures exist in a one-to-one relationship while areas of similar figures exist in a squared-to-squared relationship.  We will use that here.  

The area formula for a regular polygon is

A=\frac{1}{2}ap where a is the apothem and p is the perimeter.  We are first asked for the area of the polygon, but it would make more sense to find the perimeter first, since we need it to find the area.

P = 5(8) so

P = 40

We are given that the area of the triangle inside that polygon is 22.022 units squared.  Knowing that the area formula for a triangle is

A=\frac{1}{2}bh we can sub in what we know and solve to find the height:

22.022=\frac{1}{2}(8)h and

22.022 = 4h so

h = 5.5055 units

It just so happens that the height of that triangle is also the apothem of the polygon, so now we have what we need to find the area of the polygon:

A=\frac{1}{2}(5.5055)(40)

which gives us an area of

A = 110.11 units squared.

Here is where we can use what we know about similar figures and the relationships between perimeters and areas.  We will set up a proportion with the smaller polygon info on top and the larger info on bottom.  We know that the larger is 3 times the smaller, so the ratio of smaller to larger is

\frac{s}{l}:\frac{1}{3}

Since perimeter is one-to-one and we know the perimeter of the smaller, we can create a proportion to solve for the perimeter of the larger:

\frac{s}{l}:\frac{1}{3}=\frac{40}{x}

Cross multiply to get that the perimeter is 120 units.  You could also have done this by knowing that if the larger is 3 times the size of the smaller, then the side measure of the larger is 24, and 24 * 5 = 120.  But we used the way we used because now we have a means to find the area of the larger since we know the area of the smaller.

Area exists in a squared-to-squared relationship of the perimeter which is one-to-one.  If the perimeter ratio is 1:3, then the area relationship is

\frac{s}{l}:\frac{1^2}{3^2} which is, simplified:

\frac{s}{l}:\frac{1}{9}

Since we know the area for the smaller, we can sub it into a proportion and cross multiply to solve for the area of the larger.

\frac{s}{l} :\frac{1}{9} =\frac{110.11}{x}

A of the larger is 990.99 units squared

5 0
3 years ago
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