I think the first one is correct. Since “three less than z squared” insinuate that the value is 3 units less than the value z^2
The easiest way to graph this equation is using the x and y intercepts. Plug zero in for x to solve for y, then plug zero in got y to solve for x.
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.
The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.
Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.
So, if we sum the first N odd numbers, we have
The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have
The second sum is simply the sum of N ones:
So, the final result is
which ends the proof.
68 should be the answer... Sorry if I'm wrong
