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2 years ago

Each container must hold exactly 1 litre of water Each container must have a minimum surface area

Mathematics

1 answer:

Arada [10]2 years ago

6 0

The containers must be spheres of radius = 6.2cm

<h3>How to minimize the surface area for the containers?</h3>

We know that the shape that minimizes the area for a fixed volume is the sphere.

Here, we want to get spheres of a volume of 1 liter. Where:

1 L = 1000 cm³

And remember that the volume of a sphere of radius R is:

$V = \frac{4}{3}*3.14*R^3$

Then we must solve:

$V = \frac{4}{3}*3.14*R^3 = 1000cm^3\\\\R =\sqrt[3]{ (1000cm^3*\frac{3}{4*3.14} )} = 6.2cm$

The containers must be spheres of radius = 6.2cm

If you want to learn more about volume:

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Solve the system using multiplication for the linear combination method.

zysi [14]

(2,3) is your answer for the equation

5 0

3 years ago

1948 Olympics

mash [69]

Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

⬆ ⬆ ⬆

Low IQR Median High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR is average of 59.3 and 59.6 = 59.45

Median is average of 58.3 and 58.3 = 58.3

Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

⬆ ⬆ ⬆

Low IQR Median High IQR

Low IQR is average of 47.53 and 47.8 = 47.665

High IQR is average of 48.04 and 47.92 = 47.98

Median is average of 47.88 and 47.84 = 47.86

Interquartile range is High IQR-Low IQR

47.98-47.665=0.315

Read more on Brainly.com - brainly.com/question/14915771#readmore

8 0

3 years ago

400 reduced by 3 times my age is 163

Blababa [14]

400 - 3x = 163

Add 3x to both sides.

400 = 163 + 3x

Subtract 163 to both sides.

237 = 3x

237/3 = 79

5 0

3 years ago

How do i find the volume of these similar solids? round to the nearest tenth HELP

FromTheMoon [43]

Answer:

The volume of the larger solid is $592.6\ mm^3$

Step-by-step explanation:

<u><em>The question is</em></u>

If these solids are similar, find the volume of the larger solid

step 1

Find the scale factor

we know that

If two solids are similar, then the ratio of its corresponding sides is proportional and this ratio is called the scale factor

Let

x ----> the height of the larger solid in mm

y ----> the height of the smaller solid in mm

z ---> the scale factor

$z=\frac{x}{y}$

we have

$x=4\ mm\\y=3\ mm$

substitute

$z=\frac{4}{3}$ ---> scale factor

step 2

Find the volume of the larger solid

we know that

If two solids are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

x ----> the volume of the larger solid in cubic millimeters

y ----> the volume of the smaller solid in in cubic millimeters

z ---> the scale factor

$z^3=\frac{x}{y}$

we have

$z=\frac{4}{3}$

$y=250\ mm^3$

substitute the values

$(\frac{4}{3})^3=\frac{x}{250}$

solve for x

$(\frac{64}{27})=\frac{x}{250}$

$x=250(\frac{64}{27})$

$x=592.6\ mm^3$

4 0

3 years ago

What number raise power 3 is 729

hram777 [196]

3^x = 729
log(3^x) = log(729)
xlog(3) = log(729)
x = log(729)/log(3)
x = 6

5 0

3 years ago