Answer:
Hello! I hope I am correct! :)
Step-by-step explanation:
Let’s first calculate the black circle & the white space.
Since there is more white space than the black circle, we already know that the white space will have more probability.
These are the steps you need to do, in order to so love this problem:
1. Find the area of both, the white space and the black circle.
2. Divide both of them by the area of the square.
3. Do all these steps to find the probability/hitting.
Part A:
Black circle: ( π *1^2) / ( π 5^2) = 1/25 = .04 or 3.14% chance.
So we can tell it’s close to zero.
Part B: 1 - .04 = 0.96 or 97%
This is close to one.
(Remember I am not saying the exact, it’s a estimate)
Brainliest would be appreciated!
Hope this helps! :)
By; BrainlyAnime
The denominator of the first term is a difference of squares, such that
4<em>a</em> ² - <em>b</em> ² = (2<em>a</em>)² - <em>b</em> ² = (2<em>a</em> - <em>b</em>) (2<em>a</em> + <em>b</em>)
So you can write the fractions as
(4<em>a</em> ² + <em>b</em> ²)/((2<em>a</em> - <em>b</em>) (2<em>a</em> + <em>b</em>)) - (2<em>a</em> - <em>b</em>)/(2<em>a</em> + <em>b</em>)
Multiply through the second fraction by 2<em>a</em> - <em>b</em> to get a common denominator:
(4<em>a</em> ² + <em>b</em> ²)/((2<em>a</em> - <em>b</em>) (2<em>a</em> + <em>b</em>)) - (2<em>a</em> - <em>b</em>)²/((2<em>a</em> + <em>b</em>) (2<em>a</em> - <em>b</em>))
((4<em>a</em> ² + <em>b</em> ²) - (2<em>a</em> - <em>b</em>)²) / ((2<em>a</em> - <em>b</em>) (2<em>a</em> + <em>b</em>))
Expand the numerator:
(4<em>a</em> ² + <em>b</em> ²) - (2<em>a</em> - <em>b</em>)²
(4<em>a</em> ² + <em>b</em> ²) - (4<em>a</em> ² - 4<em>ab</em> + <em>b</em> ²)
4<em>ab</em>
<em />
So the original expression reduces to
4<em>ab</em> / ((2<em>a</em> - <em>b</em>) (2<em>a</em> + <em>b</em>))
or
4<em>ab</em> / (4<em>a</em> ² - <em>b</em> ²)
upon condensing the denominator again.
The unknown number would be 11
738 + 164 = 902
902 / 82 = 11
Ans: 11
x^5 is your answer because you add exponents when they are not in separate paretheses.
