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figured it out! sorry for posting!

There is also an area formula for a triangle:  Area  =  ½·a·b·sin(C)

where a and b are two sides of the triangle and C  is the included side.

For this problem:  Area  =  ½·6·6·sin(30^o)  =  9

Do you really need to use "choose"? Can you form a simple premutation? 

P.S. I think you are overthinking...

:)

Awesome solution Dragan!!

Amazing solution Omi!!!!!
 

See https://web2.0calc.com/questions/an-answer-was-originally-written-as-a-photo-is-there#r4

Thanks Alan,

The sum of the coefficients of any polynomial, f(x), is obtained by setting x = 1.  Hence the sum is given by f(1), which is 32 here (the rest of the information in the question is just there to confuse you!).

Here's a starter:

See if you can take it from here.

Find the area of the portion of the circle that lies outside the triangle.

It's f(x)=(x-2)^2(x+1)(x-3) because it passes y=0 at x=-1,2,3 so the solution of f(x)=0 is -1,2,3.

Let the center of a circle be O and the midpoint of BC be M

Angle B = 45º

AM = sin( 45º) * 6

XY = 2/3 * AM            (Segment XY is a side of a square inscribed in a circle.)

Area of an inscribed square    A_s = XY²

r = sin(45º) * XY

Area of a circle     A_c = r²pi

Area of a shaded segment    A = (A_c - A_s) / 4 = pi - 2   

Here is the question and answer again:

I really do not know what I am doing here. (I have forgotten it)

However.

\(|z|=5\;\;\quad so\;\; \qquad z=5e^{\theta i}\\ |w|=2\;\;\quad so\;\; \qquad w=2e^{\alpha i}\\ |\bar w|=2\;\;\quad so\;\; \qquad \bar w=2e^{-\alpha i}\\~\\ zw=5e^{\theta i}*2e^{\alpha i}=10e^{(\theta+\alpha)i}\\ |zw|^2=100\\ \left| \frac{z}{w} \right|^2=\frac{25}{4}=6.25\\~\\ z\bar w=5e^{\theta i}*2e^{-\alpha i}=10e^{(\theta-\alpha)i}=6+8i\\ \theta-\alpha=atan(\frac{8}{6})=atan(1.\dot3)\\ z\bar w=6+8i=10e^{(atan(1.\dot 3))i} \)

 I think that is right as far as it goes but it does not answer all your questions.

LaTex:

|z|=5\;\;\quad so\;\; \qquad z=5e^{\theta i}\\
|w|=2\;\;\quad so\;\; \qquad w=2e^{\alpha i}\\
|\bar w|=2\;\;\quad so\;\; \qquad \bar w=2e^{-\alpha i}\\~\\
zw=5e^{\theta i}*2e^{\alpha i}=10e^{(\theta+\alpha)i}\\
|zw|^2=100\\
\left| \frac{z}{w} \right|^2=\frac{25}{4}=6.25\\~\\
z\bar w=5e^{\theta i}*2e^{-\alpha i}=10e^{(\theta-\alpha)i}=6+8i\\
\theta-\alpha=atan(\frac{8}{6})=atan(1.\dot3)\\
z\bar w=6+8i=10e^{(atan(1.\dot 3))i}

just use the area of pyramid

Anyways, i THINK it's 

~370.85

probably wrong

ahh yes

i did it a different way

this is easier though

substitute

13(y+6)-6y=22

expand

13y+78-6y=22

isolate

13y-6y=22-78

combine

divide

done.

but you do it yourself

Which of the choices have roots (i.e.  are equal to zero) at x = -1   2   and 3   ?

   that makes it pretty easy......

no idea

all i see is 34choose34 = 1

Should be a, if i am not mistaken

f(x)=(x-2)^2(x+1)(x-3)

6*cos75º = 1.55291427

1.55291427 * 5.7955495 = 9

Isnt this and the other thing you posted,

   https://web2.0calc.com/questions/help_37869

the same question?

I know this problem.

We just had this writing last week.

They're probably in my class, even.

I'll try to find them out.