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Step2247 [10]
2 years ago
14

How were shapes named?

Mathematics
2 answers:
Vinvika [58]2 years ago
6 0

Answer:

Most shapes get their name based on the number of sides they have. Shapes also get their names from many different languages. Sometimes shapes don't have to get their name from the number of sides they have. They can get their name out of what they look like or what they smell like.

Step-by-step explanation:

PilotLPTM [1.2K]2 years ago
4 0

Answer [5]

<em>Hi, I'm Lilo, and I'm here to help you!</em>

<em>Most shapes get their name </em><em>based on the number of sides they have</em>.

<em>Maybe the "More Info" section will help.</em>

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`

More info:

<em>Most shapes get their name based on the number of sides they have. Shapes also get their names from many different languages.</em>

<em />

<em>Sometimes shapes dont have to get their name from the number of sides they have. They can get their name out of what they look like or what they smell like.</em>

<em />

<em>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~</em>

<em>Last messages:</em>

<em></em>

<em>I hope this helps :D</em>

<em>Have a great morning/afternoon/night!</em>

<em />

<u><em>- [Lilo2was2lost]</em></u>

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Step-by-step explanation:

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Find the dimensions of the rectangle with largest area that can be inscribed in an equilateral triangle with sides of 1 unit, if
prohojiy [21]
<span>Maximum area = sqrt(3)/8 Let's first express the width of the triangle as a function of it's height. If you draw an equilateral triangle, then a rectangle using one of the triangles edges as the base, you'll see that there's 4 regions created. They are the rectangle, a smaller equilateral triangle above the rectangle, and 2 right triangles with one leg being the height of the rectangle and the other 2 angles being 30 and 60 degrees. Let's call the short leg of that triangle b. And that makes the width of the rectangle equal to 1 minus twice b. So we have w = 1 - 2b b = h/sqrt(3) So w = 1 - 2*h/sqrt(3) The area of the rectangle is A = hw A = h(1 - 2*h/sqrt(3)) A = h*1 - h*2*h/sqrt(3) A = h - 2h^2/sqrt(3) We now have a quadratic equation where A = -2/sqrt(3), b = 1, and c=0. We can solve the problem by using a bit of calculus and calculating the first derivative, then solving for 0. But since this is a simple quadratic, we could also take advantage that a parabola is symmetrical and that the maximum value will be the midpoint between it's roots. So let's use the quadratic formula and solve it that way. The 2 roots are 0, and 1.5/sqrt(3). The midpoint is (0 + 1.5/sqrt(3))/2 = 1.5/sqrt(3) / 2 = 0.75/sqrt(3) So the desired height is 0.75/sqrt(3). Now let's calculate the width: w = 1 - 2*h/sqrt(3) w = 1 - 2* 0.75/sqrt(3) /sqrt(3) w = 1 - 2* 0.75/3 w = 1 - 1.5/3 w = 1 - 0.5 w = 0.5 The area is A = hw A = 0.75/sqrt(3) * 0.5 A = 0.375/sqrt(3) Now as I said earlier, we could use the first derivative. Let's do that as well and see what happens. A = h - 2h^2/sqrt(3) A' = 1h^0 - 4h/sqrt(3) A' = 1 - 4h/sqrt(3) Now solve for 0. A' = 1 - 4h/sqrt(3) 0 = 1 - 4h/sqrt(3) 4h/sqrt(3) = 1 4h = sqrt(3) h = sqrt(3)/4 w = 1 - 2*(sqrt(3)/4)/sqrt(3) w = 1 - 2/4 w = 1 -1/2 w = 1/2 A = wh A = 1/2 * sqrt(3)/4 A = sqrt(3)/8 And the other method got us 0.375/sqrt(3). Are they the same? Let's see. 0.375/sqrt(3) Multiply top and bottom by sqrt(3) 0.375*sqrt(3)/3 Multiply top and bottom by 8 3*sqrt(3)/24 Divide top and bottom by 3 sqrt(3)/8 Yep, they're the same. And since sqrt(3)/8 looks so much nicer than 0.375/sqrt(3), let's use that as the answer.</span>
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Suppose that you need to create a list of n values that have a specific known mean. Some of the n values can be freely selected.
Leno4ka [110]

As the problem indicates, degrees of freedom are the number of values ​​that can be independently selected before it is necessary to choose specific values ​​to arrive at the desired result.

The average, on the other hand, results from the sum of a list of values ​​divided by the amount of values ​​in the summed list.

Assume that the mean sought is x and consider that the list is composed of a single element a, in that case no random number can be selected, since the mean x must correspond to that number.

If the list were composed of two elements a and b, one of the two values ​​could be chosen randomly, and according to the chosen value the second should be the one whose sum with the previous one results in 2x, this given that the formula of the average \sum\limits_{i=1}^n \frac{ x_{i}}{n}.

With three values ​​a, b and c, it is possible to select two freely, since the thirteen must be the one that balances the sum of a+b, that is (a + b) + c = 3x.

Thus, in general, with n values, it is possible to select n-1 values ​​freely whose sum must be balanced by the last value so that the whole sum is nx.

Answer

In a list of \bf{n} values ​​you can assign \bf{n-1} values ​​freely, that is, you have n-1<em>degrees of freedom.</em>

6 0
3 years ago
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