Answer:
24 people.
Step-by-step explanation: If each group has to have 8 or 12 groups 24 is the smallest common multiple for both numbers.
so, no breaks on the denominator thus, the LCD will be all three then.
![\bf \cfrac{3}{x-2}+\cfrac{4}{x-1}-\cfrac{1}{x}\implies \cfrac{[3x(x-1)]+[4x(x-2)]-[(x-2)(x-1)]}{x(x-2)(x-1)} \\\\\\ \cfrac{[3x^2-3x]+[4x^2-8x]-[x^2-3x+2]}{x(x-2)(x-1)} \\\\\\ \cfrac{3x^2-3x+4x^2-8x-x^2+3x-2}{x(x-2)(x-1)} \\\\\\ \cfrac{6x^2-8x-2}{x(x-2)(x-1)}\implies \cfrac{2(x^2-4x-1)}{x(x-2)(x-1)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B3%7D%7Bx-2%7D%2B%5Ccfrac%7B4%7D%7Bx-1%7D-%5Ccfrac%7B1%7D%7Bx%7D%5Cimplies%20%5Ccfrac%7B%5B3x%28x-1%29%5D%2B%5B4x%28x-2%29%5D-%5B%28x-2%29%28x-1%29%5D%7D%7Bx%28x-2%29%28x-1%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B%5B3x%5E2-3x%5D%2B%5B4x%5E2-8x%5D-%5Bx%5E2-3x%2B2%5D%7D%7Bx%28x-2%29%28x-1%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B3x%5E2-3x%2B4x%5E2-8x-x%5E2%2B3x-2%7D%7Bx%28x-2%29%28x-1%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7B6x%5E2-8x-2%7D%7Bx%28x-2%29%28x-1%29%7D%5Cimplies%20%5Ccfrac%7B2%28x%5E2-4x-1%29%7D%7Bx%28x-2%29%28x-1%29%7D)
In this problem we need to find the value of a and b. So given that t<span>he function should be in the form f(n) = an + b and we know each value of n, then out goal is to find a and b.
For getting this purpose, we need to find a system of two equations (given that we have two unknown variables)
Therefore:
(1) f(0) = a(1) + b = 18
</span>∴ a + b = 18
<span>
(2) f(1) = a(2) + b = 24
</span>∴ 2a + b = 24<span>
Solving for a and b we have:
a = 6
b = 12
Finally:
f(n) = 6n + 12</span>
Answer:
180
Step-by-step explanation:
<em>Here,</em>
<em>b=15</em>
<em>c=9</em>
<em>Now,</em>
<em />
