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alexira [117]
3 years ago
8

In how many ways can we rearrange the digits in the number 12321? With formula and not just listing the numbers.

Mathematics
1 answer:
Archy [21]3 years ago
7 0

Answer:

With a 5-digit number, you have 5 choices for the first digit. There will be 4 digits left so you have 4 choices for the second digit. After you have chosen the first 2 digits there will be 3 remaining so you will have 3 choices for the third digit. Hence there are 2 choices for the fourth digit and only 1 choice for the fifth digit. Thus you have made 5 × 4 × 3 × 2 1 = 120 choices and there are 120 possible 5 digit numbers made from 1, 2, 3, 2 and 1

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Find the length of each side of the triangle determined by the three points P1,P2, and P3. State whether the triangle is an isos
Tanya [424]

Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$ P_{1} = (- 1, 4) $

$ P_{2} = (6, 2) $    and

$ P_{3} = (4, - 5) $

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $ (x_1, y_1) $ and $ (x_2, y_2) $ is:

                                 $ \sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+}   \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}} $

Length between $ P_1 $ and $ P_{2} $ , (Side 1) :

$ (x_1, y_1) = (- 1, 4) $     and

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2 $

$ = \sqrt{7^2 + 2^2} $

$ = \sqrt{49 + 4} $

$ = \sqrt{\textbf{53}} $

Length of Side 1 = $  \sqrt{\textbf{53}} $ units.

Distance between $ P_1 $ and P_2 , (Side 2):

$ (x_1, y_1) = (-1, 4) $

$ (x_2, y_2) = (4, - 5) $

Distance = $ \sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2 $

$ = \sqrt{ 25 + 81 } $

$ = \sqrt{\textbf{106}} $

Length of Side 2 = $ \sqrt{\textbf{106}} $ units.

Distance between $ P_2 $ and $ P_3 $ , Side 3 :

$ (x_1, y_1) = (4, 5) $

$ (x_2, y_2) = (6, 2) $

Distance = $ \sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2} $

$ = \sqrt{49 + 4} $

$ = \sqrt{\textbf{53} $

Length of Side 3  = $ \sqrt{\textbf{53}} $ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$ \bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2 $

i.e, 53 + 53  = 106

Hence, the triangle is a right - angled triangle as well.

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