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skad [1K]
3 years ago
5

A section of guttering can hold up to 20 litres of rainwater. The diameter is 11cm. Work out the length in metres.

Mathematics
1 answer:
m_a_m_a [10]3 years ago
3 0

Answer:

2.1 m

Step-by-step explanation:

The gutter is in the shape of a cylinder.

Its volume is 20 liters and its diameter is 11 cm (0.11 m).

Its radius is half of its diameter => r = 0.055 m.

The volume in cubic metres:

1 litre = 0.001 cubic metres

20 litres = 20 * 0.001 = 0.02 cubic metres

The volume of a section of guttering is given as:

V = \pi r^2h

where h = length of the guttering

Therefore:

0.02 = \pi * 0.055^2 * h\\\\0.02 = 0.009503h\\\\h = 0.02 / 0.009503\\\\h = 2.10 m

The length of the section of guttering is 2.1 m long.

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If ISO 400, SS 30, f/5.6 overexposes a picture three stops, what are the correct settings for a good exposure?
Reika [66]

Answer:

  ISO 1600

Step-by-step explanation:

"Three stops" means the picture is overexposed by a factor of 2^3 = 8. Each f-stop increase halves the exposure, so increasing from 5.6 through 8 to 11 means you have cut the exposure by a factor of 4 with the f-stop change.

Each halving of the shutter-open time also cuts the exposure in half, so reducing the shutter speed from 1/30 to 1/250 cuts the exposure by a factor of about 8.

The adjustments of shutter speed and f-stop have decreased the exposure by a factor of 4·8 = 32, which is a factor of 4 more than the factor of 8 you needed. (The shutter speed change by itself would have been sufficient.)

So, to compensate for the much smaller exposure, you need faster film, by a factor of 4. The film needs to go from ISO 400 to ISO 1600.

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4 years ago
Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match the system of equations with their so
allochka39001 [22]

Answer:

Part 1) solution {(-5,8),(3,0)} ----> y+12=x^{2}+x and x+y=3

Part 2) No solutions ----> y-15=x^{2}+4x and x-y=1

Part 3) solution [(-2,5),(3,-5)] ---> y+5=x^{2}-3x and 2x+y=1

Part 4) No solutions ---> y-6=x^{2}-3x and x+2y=2

Part 5) solution [(2,3),(8,9)] --->y-17=x^{2}-9x and -x+y=1

Part 6) solution [(-2,3),(7,-6)] --->y-15=-x^{2}+4x and x+y=1

Step-by-step explanation:

Part 1) we have

y+12=x^{2}+x ----> equation A

x+y=3 ----> equation B

Solve by graphing    

Remember that the solution of the system of equations is the intersection point both graphs

The solution are the points (-5,8) and (3,0)  

see the attached figure N 1

Part 2) we have

y-15=x^{2}+4x ----> equation A

x-y=1 ----> equation B

Solve by graphing    

Remember that the solution of the system of equations is the intersection point both graphs

The system has no solutions, because there is no point of intersection between the two graphs  

see the attached figure N 2

Part 3) we have

y+5=x^{2}-3x ----> equation A

2x+y=1 ----> equation B

Solve by graphing    

Remember that the solution of the system of equations is the intersection point both graphs

The solution are the points (-2,5) and (3,-5)  

see the attached figure N 3

Part 4) we have

y-6=x^{2}-3x ----> equation A

x+2y=2 ----> equation B

Solve by graphing    

Remember that the solution of the system of equations is the intersection point both graphs

The system has no solutions, because there is no point of intersection between the two graphs    

see the attached figure N 4

Part 5) we have

y-17=x^{2}-9x ----> equation A

-x+y=1 ----> equation B

Solve by graphing    

Remember that the solution of the system of equations is the intersection point both graphs

The solution are the points (2,3) and (8,9)  

see the attached figure N 5

Part 6) we have

y-15=-x^{2}+4x ----> equation A

x+y=1 ----> equation B

Solve by graphing    

Remember that the solution of the system of equations is the intersection point both graphs

The solution are the points (-2,3) and (7,-6)  

see the attached figure N 6

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3 years ago
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Answer:

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