Solution:
This problem is a permutation because the order matters here. This means that choosing A as King, B as Knight, C as Bishop and D as Rook results in a different arrangement from B as King, A as Knight, D as Bishop and C as Rook. We would count them both because in the first case A is King, but in the second case A is Knight.
Therefore, the possible number of ways are given below:
![19P4=\frac{19!}{(19-4)!} =\frac{19 \times 18 \times 17 \times 16 \times 15!}{15!} =19 \times 18 \times 17 \times 16 =93024](https://tex.z-dn.net/?f=19P4%3D%5Cfrac%7B19%21%7D%7B%2819-4%29%21%7D%20%3D%5Cfrac%7B19%20%5Ctimes%2018%20%5Ctimes%2017%20%5Ctimes%2016%20%5Ctimes%2015%21%7D%7B15%21%7D%20%3D19%20%5Ctimes%2018%20%5Ctimes%2017%20%5Ctimes%2016%20%3D93024)
Hence there will be 93024 ways 19 members of a chess club fill the offices of King, Knight, Bishop, and Rook.
To solve this problem, we just need to set up a simple equation. We have that angles 1 and 2 add up to equal a right angle, or 90°, and that m<2 is 35°, so we just have to do a bit of subtraction.
<1 + <2 = 90 Given
<1 + 35 = 90 Substitute 35 for <2
<1 = 55° Subtract 35 from both sides to get rid of it, since subtraction is the opposite of addition.
Therefore, m<1 = 55°.
Hope this helps!
A radius perpendicular to a chord bisects the chord, so the answer is twice 4.5
Answer: C. 9
Answer:
The last one: 70/100 = x/80
And 56 free throws.
Explanation:
70% is 70 out of 100. And if the player is free throwing a total of 80 times, then 80 is 100% of the free throws. Both of the values that represent 100% are the denominators in the fractions.
![\frac{70}{100} = \frac{x}{80}](https://tex.z-dn.net/?f=%5Cfrac%7B70%7D%7B100%7D%20%3D%20%5Cfrac%7Bx%7D%7B80%7D)
The equations are equal, and whatever you do to the bottom, you have to do to the top. 100 was multiplied by 4/5 to get 80. So 70 is multiplied by 4/5 to get x.
4/5 * 70 is 56