Cosine Law:
c² = a² + b² - 2 a b cos C
Answer:
C ) area of square c² = area of square a² + area of square b²
- area of defect 1 - area of defect 2
To factor this quadratic equation into double brackets, you need to find two numbers that add together to equal 10, but also multiply to give you 9. These two numbers can be identified as 1 and 9, which will be the numbers you place in your parentheses as follows:
(x+1)(x+9)
The probability of an event is expressed as

Given:

The probability of drwing two blue balls one after the other is expressed as

For the first draw:

For the second draw, we have only 1 blue ball left out of a total of 6 balls (since a blue ball with drawn earlier).
Thus,

The probability of drawing two blue balls one after the other is evaluted as

The probablity that none of the balls drawn is blue is evaluted as

Hence, the probablity that none of the balls drawn is blue is evaluted as
I would say y = ln x + 4
you can use a graphing calculator