Given:
The vertices of a right triangle are:

To find:
The length of the hypotenuse of the given right triangle.
Solution:
Let the vertices of the right triangle are
.
The distance formula is:

Using distance formula, we get





Similarly,




And,





Now, taking sum of squares of two smaller sides, we get




By the definition of the Pythagoras theorem, AC is the hypotenuse of the given triangle.
Therefore, the length of the hypotenuse is 10 units.
X=-2,+1,-1 ;........................,,,,,,,,,,,
Answer: 2 19/24 hours was spent in practising.
Step-by-step explanation:
During the first hour, they practiced for 5/8 of an hour. During the second hour, they practiced for 2/3 of an hour. This means that the total time for which they practiced in the first 2 hours would be
5/8 + 2/3 = 31/24 hours
During the last two hours, they first practiced for 3/5 of an hour, took a 1/2 hour break and then practiced the rest of the time. This means that the rest of the time for which they practiced is
2 - (3/5 + 1/2) = 2 - 11/10 = 9/10
Therefore, the time they spent practicing in total would be
31/24 + 3/5 + 9/10 = 67/24 =
2 19/24 hours
PEMDAS
Multiplication: 2 + 5 - 18=
Addition: 7-18=
Subtraction: -11
Answer=-11
Step-by-step explanation:
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