billy.zhang

Let's have the mother sit first. It doesn't matter where she sits cuz it's the same when rotated. Now let's seat the Chief. He has 2 spaces. Next, 6 people can be placed anywhere, so it is 6!=6*5*4*3*2*1*2=1440

h(x) has a domain of all numbers while f(x) is restricted to (-inf,7)U(7,+inf).

Hello Guest,

(a) Let's start by squaring. We have $a^2+b^2+2ab=16.$ Thus $a^2+b^2+2ab=a^4+b^4.$We soon see that the only solutions for $\boxed{ab=0.}$

(b) Problem explictly states that $a+b=4$.

(c) We know that $a+b=4$ and $ab=0$. Thus one of either $a$ or $b$ must be zero.

There are four solutions: $(a,b)=(2,0);(-2,0);(0,2);(0,-2).$

Guest, that is wrong.

We have $\sqrt{x} + 2 = \sqrt{x+100}$. Squaring both sides, we have $x+4+4\sqrt{x}=x+100$. Combining like terms, we get $4\sqrt{x}=96$, thus $\sqrt{x}=24.$ So we have $\boxed{x=24^2=576.}$ Plugging this in, we see see that it is a solution to the equation. $\square$

$\text{Add the two equations together to get }$$4(x+y+z)=2400, \text{thus } x+y+z=600.$$\text{ Because we have 3 numbers, we divide}$$\text{this by 3 to get the average:} \frac{x+y+z}{3}=\boxed{200.}$

Using the distance formula,$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2},$ we substitue them in to get $\sqrt{(2-(-4))^2+(-6-3)^2},$ which simplifies to $\sqrt{36+81}=\sqrt{117}=\boxed{3\sqrt13.}$

The 75 cancels with the -75. The 72 cancels with the -72. Since the difference between the two numbers is 3, and 3|75, all the numbers cancel out and sum to 0.

First find the value of $x.$ Expanding and combining like terms, we find that it becomes a quadratic equation, $5x^2+6x-18=0$.

Solving using the quadratic formula, we see that $x$ has two solutions- $\frac{-3\pm3\sqrt{11}}{5}$. We want the largest possible solution, so the $\pm$ becomes a $+$ instead.

Thus we have $\frac{-3+3\sqrt{11}}{5}=\frac{a+b\sqrt{c}}{d}$. Substituting these values in for each other, we find that $a=-3, b=3, c=11, d=5.$ So the value of $\frac{acd}b=\frac{-3\cdot11\cdot5}{3}=\boxed{-55}.$

Do you have any take on this yet? Have you tried this problem on your own? Since it's marked homework, you should try that first.

$\frac{343}{3}=114 \frac13$. What did they do to the dogs?!?!

Question-wise, it's $114 \frac13 \cdot 2 = \boxed{228 \frac23 \text{non-dogs.}}$ 

 Angle $\text{ETF}$ is $105$º

I think this is only half of the problem! 

However, researching, I found this problem:

N is a four-digit positive integer. Dividing N by 9 , the remainder is 5. Dividing N by 7, the remainder is 3. Dividing N by 5, the remainder is 4. What is the smallest possible value of  N?