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?