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tankabanditka [31]
3 years ago
5

A vessel without a lid has the shape of a cylinder. The height of the vessel is 1.5 m, and the diameter of the base represents 4

0% of the height. Determine if 1 kg of paint is enough to completely paint the vessel inside and out, if it is known that for 1 square meter of surface 150g of paint is consumed.​
Mathematics
1 answer:
salantis [7]3 years ago
7 0

Answer:

Yes, 1 kg of paint is enough to completely paint the inside and outside surfaces of the vessel.

Step-by-step explanation:

For the given vessel without a lid,

diameter = 0.4 x 1.5

               = 0.6 m

radius = \frac{diameter}{2} = \frac{0.6}{2}

           = 0.3 m

Total external surface area of the vessel = 2\pirh + \pir^{2}

                             = [2 x \frac{22}{7} x 0.3 x 1.5] + [\frac{22}{7} x (0.3)^{2}]

                             = 2.8286 + 0.2829

                             = 3.1112

Total external surface area of the vessel is 3.1112 square meter.

Total internal and external surface area = 2 x 3.1112

                                                              = 6.2224 m^{2}

But for 1 square meter, 150 g of paint is consumed. Thus;

for 6.2224 m^{2} = 6.2224 x 150 g

                       = 933.36 g

Thus, 6.2224 m^{2} of area would require 933.36 g of paint.

So that,

\frac{933.36}{1000} = 0.93336 kg

Therefore, 1 kg of paint is enough to completely paint the inside and outside surfaces of the vessel.

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A 24-ounce mocha beverage with whipped cream has 25% of the calories allowed on a 2,000-calorie-per-day diet. What percentage of
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gayaneshka [121]

Answer:

The correct option is;

A dilation by a scale factor of Two-fifths and then a translation of 3 units up

Step-by-step explanation:

Given that the coordinates of the vertices of square S are

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Given that the coordinates of the vertices of square S' are

(0, 1), (0, 3), (2, 3), (2, 1)

We have;

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For square S', where (x₁, y₁) = (0, 1) and (x₂, y₂) = (0, 3)

Length of side, s₂, for square S' is s₂ = √((3 - 1)² + (0 - 0)²) = √(2)² = 2

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The transformation of square S to S' involves a dilation of s₂/s₁ = 2/5

The after the dilation (about the origin),  the coordinates of S becomes;

(0, 0) transformed to (remains at) (0, 0) ....center of dilation

(5, 0) transformed to (5×2/5, 0) = (2, 0)

(5, -5) transformed to (2, -2)

(0, -5) transformed to (0, -2)

Comparison of (0, 0), (2, 0), (2, -2), (0, -2) and (0, 1), (0, 3), (2, 3), (2, 1) shows that the orientation is the same;

For (0, 0), (2, 0), (2, -2), (0, -2) we have;

(0, 0), (2, 0) the same y-values, (∴parallel to the x-axis)

(2, -2), (0, -2) the same y-values, (∴parallel to the x-axis)

For (0, 1), (0, 3), (2, 3), (2, 1) we have;

(0, 3), (2, 3) the same y-values, (∴parallel to the x-axis)

(0, 1), (2, 1) the same y-values, (∴parallel to the x-axis)

Therefore, the lowermost point closest to the y-axis in (0, 0), (2, 0), (2, -2), (0, -2) which is (0, -2) is translated to the lowermost point closest to the y-axis in (0, 1), (0, 3), (2, 3), (2, 1) which is (0, 1)

That is (0, -2) is translated to (0, 1) which shows that the translation is T((0 - 0), (1 - (-2)) = T(0, 3) or 3 units up

The correct option is therefore a dilation by a scale factor of Two-fifths and then a translation of 3 units up.

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