Answer:
choice 4) right scalene (one right angle and the side opposite that angle is the longest side)
choice 1) obtuse isosceles (one angle greater than 90 degrees and 2 equal length legs)
Step-by-step explanation:
Answer:
<em>-1,3,5,9,13</em>
Step-by-step explanation:
2(-3)+5=-6+5=-1
2(-1)+5=-2+5=3
2(0)+5=0+5=5
2(2)+5=4+5=9
2(4)+5=8+5=13
Answer:
1) f(g(2)) = 24
2) f(g(-1)) = -4
Step-by-step explanation:
1) GIven f(x) = x²+2x and g(x) = 2x
f(g(x)) = f(2x)
f(2x) = (2x)² + 2(2x)
f(2x) = 4x² + 4x
f(g(x)) = 4x² + 4x
f(g(2)) = 4(2)² + 4(2)
f(g(2)) = 16+8
f(g(2)) = 24
2) f(x) = x+1 and g(x) = 5x
f(g(x)) = f(5x)
f(5x)= 5x + 1
f(g(x)) = 5x + 1
f(g(-1)) = 5(-1) + 1
f(g(-1)) = -5+1
f(g(-1)) = -4
Answer:
There are 1% probability that the last person gets to sit in their assigned seat
Step-by-step explanation:
The probability that the last person gets to sit in their assigned seat, is the same that the probability that not one sit in this seat.
If we use the Combinatorics theory, we know that are 100! possibilities to order the first 99 passenger in the 100 seats.
LIke we one the probability that not one sit in one of the seats, we need the fraction from the total number of possible combinations, of combination that exclude the assigned seat of the last passenger. In other words the amount of combination of 99 passengers in 99 seats: 99!
Now this number of combination of the 99 passenger in the 99 sets, divide for the total number of combination in the 100 setas, is the probability that not one sit in the assigned seat of the last passenger.
P = 99!/100! = 99!/ (100 * 99!) = 1/100
There are 1% probability that the last person gets to sit in their assigned seat
Your intention with the wording "consecutive integers of 12 feet" is not clear. Shown are triangles with sides 12 and 13 (a nice Pythagorean triple) as well as 11 and 12.
