Melody

How many ways are there to divide a group of 6 friends among the basketball team, the soccer team, and the track team? (Each team could have anywhere from 0 to 6 of the friends on it. Assume the friends are distinguishable.)

Possibilities

0,0,6          1 way to seperate. 3 alternatives for teams  =3

0,1,5         6C1 way to seperate. 3! alternatives for teams  =36

0,2,4         6C2  way to seperate. 3! alternatives for teams  =36

0,3,3         6C3/2  way to seperate. 3! alternatives for teams  =18

1,1,4         6C4  way to seperate. 3! alternatives for teams  =90

1,2,3         6C3*3  way to seperate. 3! alternatives for teams  =360

2,2,2         6C2*4C2/6  way to seperate. 3! alternatives for teams  =90

3+36+36+18+90+360+90 = 633

I could easily have made a mistake.  If you get the correct answer, please share it here.

I'd start by considering factors of 24    

plus or minus  1,2,3,4,6,8,12,24     That's a lot  16 in total

Now trial and error to see which one will divide in perfectly.

\(f(x)=x⁴-2x³-13x²+38x-24\\ f(1)=1-2-13+38-24 =39-39=0\;\;great\\ (x-1)\text{ is a factor} \)

Now I would divide f(x) by  x-1    to get the dividend then start the process again.

OR you could just look for another possibility out of the list.

What is the value of b+c if x^2+bx+c>0 only when \( x\in (-\infty, 2)\cup(2,\infty)?\)?

\(x^2+bx+c>0 \)

this is a concave up parabola.

If this function is greater than zero for all values of x except for x=2  then the vertex is  (2,0)

And there will only be one root.

so

\(x^2+bx+c=(x-2)^2\\ b=-4,\;c=4\\ b+c=0 \)

The parabola and the circle should just touch.

The blue circle is your solution.  As you can see your circl does not touch the parabola.

So the parabola is always more than 3.14 units from the origin.

Your answer is not correct.

The green circle is guests solution.  As you can see, his/her cirlce does appear to be tangential to the parabola. 

So the green answer appears to be correct.

1*3*4*7 ways

Thanks Ginger.

I am coming I am having the prawns  

Technically they may be different animals but when Americans talk of eating a shrimp, that is what we call a prawn. 

3y-12 = 12-(y+6)

Now solve

It can be seen from this graph that guest answer is most likely correct.

asinus, I am sorry but your answer is not correct.  (maybe it is just a careless error, I have not checked)

You want it exact, you could have done it for yourself. 

But here it is.

\(tan\alpha=\frac{3}{4}=0.75\\ \alpha=atan0.75\\ sin(2\alpha )=\frac{3}{y}\\ y= \frac{3}{sin(2\alpha)}\\ Green Area=2*0.5*y*y*sin(180-2\alpha)\\ Green Area=y*y*sin(2\alpha)\\ Green Area=y*y*\frac{3}{y}\\ Green Area=3y \\Green Area=\frac{9}{sin(2\alpha)} \\Green Area=\frac{9}{2sin\alpha cos\alpha} \\Green Area=\frac{9}{2*\frac{3}{5}*\frac{4}{5}}\\ \\Green Area=\frac{9*25}{24}\\ \\Green Area=9.375\)

Two 3 * 4 rectangles overlap. Find the area of the overlapping region (which is shaded green).

\(tan\alpha=\frac{3}{4}=0.75\\ \alpha=atan0.75\approx 36.87^0\\ 2\alpha\approx 73.74^0\\ sin73.74=\frac{3}{y}\\ y\approx \frac{3}{sin73.74}\approx3.125\\ Green Area=2*0.5*y*y*sin(180-2\alpha)\\ Green Area=y*y*sin(2\alpha)\\ Green Area=y*y*\frac{3}{y}\\ Green Area=y*3\\ Green Area = 3*3.125\\ Green Area \approx 9.375u^2\\ \)

LaTex:

tan\alpha=\frac{3}{4}=0.75\\
\alpha=atan0.75\approx 36.87^0\\
2\alpha\approx 73.74^0\\
sin73.74=\frac{3}{y}\\
y\approx \frac{3}{sin73.74}\approx3.125\\
Green Area=2*0.5*y*y*sin(180-2\alpha)\\
Green Area=y*y*sin(2\alpha)\\
Green Area=y*y*\frac{3}{y}\\
Green Area=y*3\\
Green Area = 3*3.125\\
Green Area \approx  9.375u^2\\