Answer:
a) 66
b) 45
Step-by-step explanation:
We have 12 horses and after the first race 2 of them have to be eliminated
that condition is equal to consider that 10 horses will be available for the second race.
The number of possiblities we have for the first race is the combination of 12 horses in group of 10
C¹²₁₀ = 12!/ (2!)* (12-2)! = 12*11/2 = 66
For the second race
C¹⁰₈ = 10! / 8! * (10-8)! = 10*9/ 2 = 45
Answer:
First choice
Step-by-step explanation:
It passes through (4, 0) and (0, -4)
Slope = (0 - (-4))/(4 - 0) = 4/4 = 1
so y - 0 = 1*(x - 4)
x - y = 4
8x - 8y = 32
First choice
D = sqrt(3s^2) where s is the length of the side. Solving for s,
<span>3s^2 = d^2 iff </span>
<span>s^2 = d^2 / 3 iff </span>
<span>s = sqrt(d^2 / 3) </span>
<span>= d / sqrt(3) or d sqrt(3) / 3 </span>
<span>Surface area of the cube = 6 s^2. Thus, </span>
<span>A = 6 (d / sqrt(3))^2 </span>
<span>= 6d^2 / 3 </span>
<span>= 2d^2 </span>
<span>Volume = s^3. Thus, </span>
<span>V = (d / sqrt(3))^3 </span>
<span>= d^3 / 3sqrt(3) </span>
<span>= d^3 sqrt(3) / 9</span>
Answer:
20/9
Step-by-step explanation:
Find the GCD (or HCF) of numerator and denominator
GCD of 80 and 36 is 4
Divide both the numerator and denominator by the GCD
80 ÷ 4
36 ÷ 4
Reduced fraction: 20/9
