The first 4 on the left has a value of 400. The next 4 has a value of 40. The first 4 is in the hundreds place and the second 4 is in the 10's place.
So, 1/3 of something is 21.
1/3 of x is not the full x: we know that the x is 21 and even more. So, it must be bigger than 21, since it has 21 and something else (and since 21 is not negative): the solution MUST be greater than 21.
Answer:
5Etxxug
Step-by-step explanation:
64d64s585i is a solution of the linear
Looking at Pascal's Triangle, specifically at the row that starts with 1, 10, etc we see the value 210 in the 4th slot (start the count at 0) since 10-6 = 4
Or you can use the combination formula nCr to get the same result
nCr = (n!)/(r!*(n-r)!)
10C4 = (10!)/(4!(10-4)!)
<span>10C4 = (10!)/(4!*6!)
</span><span>10C4 = (10*9*8*7*6!)/(4!*6!)
</span><span>10C4 = (10*9*8*7)/(4!)
</span><span>10C4 = (10*9*8*7)/(4*3*2*1)
</span>10C4 = 210
Either way, the final answer is Choice C) 210
