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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​.

The answer that i got is (-inf, -15], [0.6,+inf)

Look at the diagram:

Find angle \(\alpha \) using the cosine rule    (i.e. \(cos(a) = (b^2 + c^2 - a^2)/(2bc)\) )

Take the ratio of the arc of the circle subtended by \(2\alpha \)  to that of the circumference of the whole circle to get the probability.

i.e. probability = \(2\alpha / (2\pi) \)

(Note that the lower point could be on the opposite side of the circle hence \(2\alpha \) not just \(\alpha \))

Put one adult anywhere.

then there are 2 posibilities for how to sit the other 2 adults 

Tthen 3! = 6 ways to seat the chiildren.

So that is  2*6=12 ways

What is the x-intercept of the line?

Hello tomtom!

\(m=2\\ b=2\\ y=mx+b\\ 2x+2=0\\ 2x=-2\\ \color{blue}x=-1\)

The x-intercept of the line is -1.

  !

The amoeba were actually found to be smaller, so there should be more of them, not less

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​.

3x -7   in the required interval  can only be   (2  thru   6 1/3 )

   then 2x    to be in the interval can only be  (-1/2 through 6)         so     the common factors are  x =   (2,6)      

Pick any two points on the line to calculate the slope = rise/run = 1/2 

    then    y = 1/2 x + b      sub in any point to calculate the value of 'b'

               2 = 1/2(0) + b     shows b = 2

 y = 1/2 x + 2    re-arrange

1/2 x - y = -2     multip;ly through by 2  ( to get rid of the fraction)

x - 2y = -4 

y-axis intercept occurs when x = 0 

    3(0) -6y = 12 + 2(0) - 4y 

          -2y = 12 

             y = -6                y intercept = (0,-6) 

x axis intercept occurs when y = 0 

   3x - 6(0) = 12 + 2x - 4(0) 

       x = 12                      x intercept = (12, 0 )

Using Pythagorean theorem    d^2 = (-6)^2 + 12^2 

                                                 d = sqrt (180)  = 6 sqrt (5) 

Write the equation in the form $Ax + By = C.

Hello tomtom!

P(-8,2)

Q(1,7)

\(m=\frac{y_P-x_P}{y_Q-x_Q}=\frac{2-(-8)}{7-1}=\frac{5}{3}\\ y=m(x-x_Q)+y_Q\\ y=\frac{5}{3}(x-1)+7\\ y=\frac{5}{3}x-\frac{5}{3}+7\\ -\frac{5}{3}x+y=\frac{16}{3}\ |\ \cdot (-3)\\ \color{blue}5x-3y=-16\)

  !

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​]​.

First, we can use the given functional equation to show that f(0) = 1. We have f(0) + f(0) = f(0 + 0) = f(0) = 2. This equation has two solutions: f(0) = 0 or f(0) = 1. If f(0) = 0, then f(x) = 0 for all x, since f(x) = f(x + 0) = x + f(0) = x + 0 = x. However, we know that f(0) = 2, so f(0) must be 1.

Now, we can use the given functional equation again to show that f(2012) = 2013. We have f(2012) + f(0) = f(2012 + 0) = f(2012) = 2012 + f(0) = 2012 + 1 = 2013. This equation has only one solution: f(2012) = 2013.

Therefore, the answer is 2013.

There are two cases to consider:

Case 1: One child sits next to two adults, and the other two children sit next to one adult each.

Case 2: Each child sits next to one adult.

In Case 1, we can choose which child sits next to two adults in 3 ways. Once we have chosen that child, we can seat the other two children and the three adults in a row in 6! ways. However, we have overcounted the number of arrangements by a factor of 2, since we have counted each arrangement as if the children were distinguishable.

In Case 2, we can seat the three adults in a row in 3! ways. Then, we can seat the three children between the adults in 3! ways. 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 if each child must be next to at least one adult is:

(3)(6!)/2 + (3!)(3!)/3 = 2! (6 + 3!) = \boxed{24}.

When the balls are indistinguishable, there is only one way to place them in the boxes. We can simply put one ball in each box.

However, the boxes are also indistinguishable, so there are 8! ways to arrange the boxes.

Therefore, there are 8!​ ways to place 4 balls in 8 boxes if neither the balls nor the boxes are distinguishable.