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Rzqust [24]
3 years ago
12

Question 1

Mathematics
1 answer:
Alex777 [14]3 years ago
5 0
Riki ...
Most questions on Brainly offer 5 points in return for answering one question,
but here you've listed five of them to be answered for 5 points.  That's kind of
much.
Also, it's not clear what cylinders Question-2 and Questioin-3 refer to. 
They might be the two pictures under the question, but we don't rally know.
Fortunately, all five of your questions are answered with the same formula ...
the formula for the volume of a cylinder.
Here it is:

         Volume of a cylinder = ( π ) · (radius)² · (height or length)  .

                                       V  =  π  R²  H  .

That's it.  That's all you need to answer any or all of the five questions.
So I've actually given you everything you need to answer them, and in
doing that, I've already earned 5 points.  But, being the ever helpful guy
that I am, I'll work out two of the questions for you, to show you how to
use the formula.

Question #1).  <span>A cylinder has a radius of 14 m and a height of 6 m.

              </span>Volume  =  π  R²  H 

                           =  (π)  (14m)²  (6m)

                           =  (π)  (196m²)  (6m)

                           =      1,176 π  m³  =  71,763,930 π in³

(Notice how confusingly and sloppily the question is written . . .
the dimensions of the cylinder are given in meters, but the
answer choices are listed in cubic-inches.  I've shown the
answer in both units.)

Question 5).  <span>A can has a radius of 2.8 feet and holds 86.21 cubic feet.

           </span><span>Volume  =  π  R²  H

           86.2 ft³  =  π (2.8 ft)² (Height)

            86.2 ft³  =  π (7.84 ft²) (Height)

           Height  =  (86.2 ft³) / (7.84 π ft²)

                        =         3.5 ft

</span>
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On a linear X temperature scale, water freezes at −115.0°X and boils at 325.0°X. On a linear Y temperature scale, water freezes
belka [17]

Answer:

The current temperature on the X scale is 1150 °X.

Step-by-step explanation:

Let is determine first the ratio of change in X linear temperature scale to change in Y linear temperature scale:

r = \frac{\Delta T_{X}}{\Delta T_{Y}}

r = \frac{325\,^{\circ}X-(-115\,^{\circ}X)}{-25\,^{\circ}Y - (-65.00\,^{\circ}Y)}

r = 11\,\frac{^{\circ}X}{^{\circ}Y}

The difference between current temperature in Y linear scale with respect to freezing point is:

\Delta T_{Y} = 50\,^{\circ}Y - (-65\,^{\circ}Y)

\Delta T_{Y} = 115\,^{\circ}Y

The change in X linear scale is:

\Delta T_{X} = r\cdot \Delta T_{Y}

\Delta T_{X} = \left(11\,\frac{^{\circ}X}{^{\circ}Y} \right)\cdot (115\,^{\circ}Y)

\Delta T_{X} = 1265\,^{\circ}X

Lastly, the current temperature on the X scale is:

T_{X} = -115\,^{\circ}X + 1265\,^{\circ}X

T_{X} = 1150\,^{\circ}X

The current temperature on the X scale is 1150 °X.

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