Assuming the order does not matter, you want the number of combinations of 9 things taken 5 at a time. The combinations can be shown as C(9,5), 9C5.
C(9, 5) =
9/5(9-5) =
9*8*7*6*5 / 5*4
The 5 terms cancel.
9*8*7*6 / 4*3*2 =
9*7*2 =
126
The above change is because 4*2 cancels the 8 in the numerator and 6/3 = 2
Therefore, the solution is 126.
Answer:
option 4.
16 square units
Step-by-step explanation:
as we do not have the measures of the sides, but if the points of the vertices with Pythagoras we can calculate the sides.
P = (2 , 4)
S = (4 , 2)
we have to subtract the values of p from s
PS = (4 - 2 , 2 - 4)
PS = (2 , -2)
by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2
h: hypotenuse
c1: leg 1
c2: leg 2
PS^2 = 2^2 + -2^2
PS = √ 4 + 4
PS = √8
PS = 2√2
S = (4 , 2)
R = (8 , 6)
SR = (8-4 , 6-2)
SR = (4 , 4)
by pitagoras h ^ 2 = c1 ^ 2 + c2 ^ 2
h: hypotenuse
c1: leg 1
c2: leg 2
SR^2 = 4^2 + 4^2
SR = √ (16 + 16)
SR = √32
SR = 4√2
having the values of 2 of its sides we multiply them and obtain their area
PS * RS = Area
2√2 * 4√2 =
16
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 = 40320.
Answer:
60 square units
Step-by-step explanation:
