It's $\frac{51}{40}$, greater than $1$ but less than $1 \frac{1}{2}$.
Answer is R.
$a+b+c=750$
$a-30, b+30, c$
$a-30, b-20, c+50$
$a-30=b-20=c+50$
Solve.
$c=j+20$
$2j+20=100$
$j=40$
$c=60$
$4(40-t)=60$
Find $t$.
There is no diagram!!!
The answer is $\infty$.
To do this, use the code snippet:
if (4){
counter++;
}
The counter will always increase with whatever bounds you have set.
$P(G(a))=P(4+8a)=4+2\sqrt{6+8a}$.
We must have $8a+6\ge 0\implies 8a\ge -6\implies a\ge -\frac{3}{4}$. The minimum value of $a$ is $-\frac{3}{4}$, the maximum is non-existent.
All positive integers satisfy the statement "4".
(I'm not joking, write code that says )
$k^6\equiv 4\pmod{10}$ has solutions $k\equiv \pm 2\pmod{10}$.
So $2,8,12,18,22,28, ...$ all work.
10 of the levels is 1/3 of the total.
30 levels in total.
