Yeetww

A=2/3=50/75

B=1536

C=8/25=24/75

D=-16

E=$3\sqrt{2}$$\approx4.24264068$

So the order is

$C,A,E,D,B$

Compute $\dbinom53+\dbinom63+\dbinom73$.

Otherwise there would either be zero solutions or infinite.

Sorry, I am unable to solve your problem. For two variable equations to be solved, there must be two equations.

Since the average is small, we can safely assume that it is easy to find using trial and error.

Row 0's average is 1.

Row 1's average is 1.

Row 2's average is $\dfrac43$

Row 3's average is $\dfrac84$ which is simply 2.

We are not done yet, though.

Row three consists of $1,3,3,1$,

so thus $\boxed{n=3}$

I downvoted two times

Lets see if my downvotes work.

convert first into $b=14-a$

then insert into second:

$a^3+(14-a)^3=812+a^2+(14-a)^2$

the end result is

$a=\dfrac{35\pm2\sqrt{35}}{5}$

insert a into first equation:

$b=\dfrac{35\pm2\sqrt{35}}{5}$

so $\boxed{a,b=\dfrac{35\pm2\sqrt{35}}{5}}$

$A=\dfrac46=\dfrac23=0.\overline6\\ B=1535\\ C=\dfrac{8}{25}=0.32\\ D=2\cdot-8=-16\\ E=3\cdot \sqrt[2]{2}=3\sqrt[2]2=4.24...$

So the order is d,c,a,e,b.

Solve both equations:

$4x+2<-14 \rightarrow 4x<-16 \rightarrow \boxed{x<-4}$

Next one:

$3x-5>16 \rightarrow 3x>21\rightarrow \boxed{x>7}$

This means that the correct answer is B.

Use a calculator... $\dfrac{733}{125}=5.864$

Or just use trial and error.. $(5\cdot125\rightarrow 625)<733<(6\cdot125\rightarrow750)$

It means that the answer should be 5, as six is too much and 5 is just right.

a more proper explanation:

d is the answer because the x ranges from 1 to 4, and since they all vary by one, you should choose d.

Solution:

(a) The number of ordered triples $(a,b,c),$ where $1 \le a, b,c \le n,$ is $n^3$ We can classify these triples into three categories.

Case 1:  $a,b,c$ are all equal.

There are $n$ ordered triples in this case.

Case 2: Two of $a,b,c$ are equal, and the third is different.

There are $n$ ways to choose the two equal values, then $n-1$ ways to choose the third value that is different. There are then $3$ ways to arrange the numbers within the ordered triple, so the number of ordered triples in this case is $3n(n-1).$

Case 3: $a,b,c$ are all different.

There are $\dbinom{n}{3}$ ways to choose three different values. There are then $3!=6$ ways to arrange them, so the number of ordered triples in this case is $6\dbinom{n}{3}$

By counting the number of triples $(a,b,c)$ in two different ways, we arrive at the conclusion 

$n^3=n+3n(n-1)+6\dbinom{n}{3}.$

(b) The number of subsets of $\{1,2,3,...,n+2\}$ containing three different numbers is $\dbinom{n+2}{3}$

We can also count the number of subsets as follows. Let the subset be $\{a,b,c\},$ where \(a  Then the middle element  $b$ is at least $2$ and at most  $n+1$ so we can let  $b=k+1$ where  $1\le k \le n.$

Once the middle element  $b=k+1$ is chosen, the smallest element  $a$ must be between $1$ and  $k$ inclusive, so there are  $k$ ways of choosing the smallest element  $a$. The largest element  $c$ must be between  $k+2$ and  $n+2$ inclusive, so there are  $(n+2)-(k-2)+1$ ways of choosing the largest element  $c$. So the number of subsets where the middle element is  $k+1$ is  $k(n-k+1)$ Summing over  $1 \le k \le n$ we get

$\dbinom{n+2}{3}=(1)(n)+(2)(n-1)+(3)(n-2)+\cdots+(k)(n-k+1)+\cdots+(n)(1).$
 

Let us call the circle's center $O$. We first note that if $A$ and $B$ are points on the circle, then triangle $AOB$ is isosceles with $AO=BO$. Therefore, if $AOB$ is an obtuse triangle, then the obtuse angle must be at $O$. So $AOB$ is an obtuse triangle if and only if minor arc $AB$ has measure of more than  $\dfrac{\pi}{2}$$(90^{\circ})$.

Now, let the three randomly chosen points be $A_0$,$A_1$, and $A_2$. Let $\theta$ be the measure of minor arc $A_0A_1$. Since $\theta$ is equally likely to be any value from $0$ to $\pi$, the probability that it is less than $\dfrac{\pi}{2}$ is $\dfrac{1}{2}$.

Now suppose that $\theta<\dfrac{\pi}{2}$. For the problem's condition to hold, it is necessary and sufficient for point $A_2$ to lie within $\dfrac{\pi}{2}$ of both $A_0$ and $A_0$ along the circumference. As the diagram below shows, this is the same as saying that $A_2$ must lie along a particular arc of measure $\pi-\theta$.



The probability of this occurrence is $\dfrac{\pi-\theta}{2\pi}=\dfrac{1}{2}-\dfrac{\theta}{2\pi}$, since $A_2$ is equally likely to go anywhere on the circle. Since the average value of $\theta$ between $0$ and $\dfrac{\pi}{2}$ is $\dfrac{\pi}{4}$, it follows that the overall probability for $\theta<\dfrac{\pi}{2}$ is $\dfrac{1}{2}-\dfrac{\pi/4}{2\pi}=\dfrac{3}{8}$.

Since the probability that $\theta<\dfrac{\pi}{2}$ is 1/2, our final probability is $\dfrac{1}{2}\cdot\dfrac{3}{8}=\dfrac{3}{16}$.