bingboy

Let the row of Pascal's triangle be [1, n, \binom{n}{2}, \dots, \binom{n}{2}, n, 1.]Then the average of the entries in this row is [\frac{1 + n + \binom{n}{2} + \dots + \binom{n}{2} + n + 1}{n + 1} = \frac{1 + 2(1 + \binom{n}{2}) + 1}{n + 1} = \frac{4 \binom{n}{2} + 3}{n + 1}.]Since the average is $2$, we have [\frac{4 \binom{n}{2} + 3}{n + 1} = 2,]so 4(2n​)+3=2(n+1). Solving for $n$, we get n=6​.

We can solve the first equation for s, getting s=3/2​−5t. Substituting this into the second equation, we get: 3t−6((3/2​−5t))=9⇒−27t+9=−9⇒ t = 2/3.

Substituting this back into the equation for s, we get s = 3/2−5(2/3​) = 1/6

Therefore, the solution is (s,t) = (1/6,2/3).

Simplifying the given inequalities, we find [-7x \le 20 \qquad \text{and} \qquad -4x \le 36.]Both inequalities are satisfied for x≥−5, so the answer is x≥−5​.

We simplify the inequalities to find [7x \le -14, \qquad 4x \ge -10.]Both inequalities are satisfied by x≤−2, so the answer is x≤−2​.

Because △ABC is a right triangle, it follows that the circumcenter of △ABC is the midpoint of the hypotenuse. Thus we have OA=OB=OC=13. By Pythagoras, we also have AC2+BC2=AB2, so 21​(AB+BC)(AB−BC)=AB2, so [AB = \frac{2AC^2}{AB + BC} = \frac{2 \cdot 24^2}{24 + 10} = \boxed{36}.]

The radius of the incircle of △ABC is [r = \frac{K}{s} = \frac{2(24)(10)(26)}{(24 + 10 + 26)} = 2.]Then the area of the incircle is K=πr2=4π​, so x−y=362π−4π=36π(36−1)​=1246π​.

To seat three children at a circular table such that each child must be next to two adults, we can create a sequence of three adults and three children in which each child is in between two adults. There are 6! ways to arrange the adults and children in a sequence. However, we have overcounted the number of arrangements by a factor of 3, since we have counted each arrangement as if the children were distinguishable.

Therefore, the total number of ways to seat the three adults and three children at a circular table such that each child must be next to two adults is:

6!/3 = \boxed{240}.

The diameter of the amoeba is given in micrometers, which is the same as microliters (μL). The volume of a sphere is proportional to the cube of its radius, so the volume of an amoeba is proportional to the cube of its diameter.

If the diameter of the amoeba is halved, the volume is reduced by a factor of 23=8. Therefore, the new estimate of the number of amoeba in a slug is 100,000/8=12,500​.

Simplifying, we have −4z≤−9 and 7z≤11. These yield z≥49​​ and z≤711​​, which is impossible since 711​<49​. Hence, there are no solutions​.

Simplifying the given inequalities, we have:

14 \le 9w \le 21,

which implies:

\frac{14}{9} \le w \le \frac{21}{9}.

Therefore, the solution is [14/9,21/9​]​.