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Lina20
2 months ago
13

Which classification best represents a triangle with side lengths 6 cm, 10 cm, and 12 cm? acute, because 62 + 102 < 122 acute

, because 6 + 10 > 12 obtuse, because 62 + 102 < 122 obtuse, because 6 + 10 > 12
Mathematics
2 answers:
lawyer [12.5K]2 months ago
7 0
The Pythagorean theorem asserts that
the sum of the squares of the two shorter sides (legs) of a right triangle equals the square of the longest side.

A related principle from this theorem is useful to address this issue:
If the total of the squares of the shorter sides of a triangle surpasses the square of the longest side, the angle in question is acute...... (case 1)
On the other hand, if the total of the squares of the shorter sides is less than that of the longest side, the triangle is obtuse......(case 2)

In this case
6^2+10^2 = 36+100=136 <12^2=144
Thus, this is case 2, indicating that the triangle is obtuse.
PIT_PIT [12.4K]2 months ago
3 0

The classification is C, xotwod

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In 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams.
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The world’s largest pumpkin weighed 1,805 kilograms.
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Very large or very small numbers are best input with exponential notation. For instance, the number 8,000,000 (eight million) is
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9*10^{12}km

Step-by-step explanation:

9,460,730,472,580.8 km

To calculate 5% of our number, we start by multiplying by 0.05

9,460,730,472,580.8 * 0.05 = 473,036,523,629

Next, by adding and subtracting this 5% from our original number, we can find the minimum and maximum possible values of our final answer in the specified range.

9,460,730,472,580.8 + 473,036,523,629 = 9.933767*10^{12}\\9,460,730,472,580.8 - 473,036,523,629 =8.98769395*10^{12}

We can ascertain that 9*10^{12} fits within the acceptable range, as:

8.9*10^{12} \leq 9*10^{12} \leq 9.9 * 10^{12}

Therefore, our final answer will be:

9*10^{12}km

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A 20-person fraternity is making their weekend plans around 5 different parties in IV. Each brother will attend exactly one of t
Leona [12618]

Answer:

3876

Step-by-step explanation:

We have the following data:

Members of the fraternity = 20

Their attendance at five parties is organized into groups of four.

This indicates that each group consists of 4 individuals.

At least one brother will be present at exactly one of the gatherings. (The brothers are indistinguishable).

Thus, precisely one brother at a gathering leaves (20 - 1) = 19, due to their indistinct nature.

Group sizes are in fours.

Calculating 19C4:

From: nCr = n! /(n-r)! r!

19C4 = 19! / (19 - 4)! 4!

= 19! / 15! 4!

= (19 * 18 * 17 * 16) / (4 * 3 * 2)

= 93024 / 24

= 3876

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2 months ago
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