X^2 - 8x-9=0
D = b^2- 4*a*c = 64- 4* (-9) = 64+36=100 =10^2
X1 = (8-10)/2 = -1
X2= (8+10)/2= 9
Answer: The loser's card shows 6.
Explanation: Let's start by naming the first student A and the second student B.
Since the product of A and B are either 12, 15, or 18, let's list every single possibility, the first number being A's number and the second number being B's number.
1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
4 3
5 3
6 2
6 3
9 2
12 1
15 1
18 1
Now, the information says that A doesn't know what B has, so we can immediately cross off all of the combinations that have the integer appearing once and once ONLY off, because if it happened once only, A would know of it straight away. Now, our sample space becomes much smaller.
1 12
1 15
1 18
2 6
2 9
3 4
3 5
3 6
6 2
6 3
Using this same logic, we know that we can cross off all of the digits that occur only once in B's column.
2 6
3 6
Now, A definitely knows what number B has because there is only one number left in B. Hence, we can conclude that the loser, B, has the integer 6.
3x-10>4+7x-3
3x-10>1+7x
-4x-10>1
-4x>11
x<-11/4
Hope Your Thanksgiving Goes Well, Here's A Turkey
-TheKoolKid1O1
