Melody

x^2 - 60x + 851

factor(851) = 23*37

23+37=60

(x+23)(x+37)

a=37 b=23

You can finish it.

I think you did that brilliantly.   

I used your method and got the same answer. 

Neither digit is zero, 2 of each one

9C2*4!/(2!2!) = 9C2* 6 = 36*6 = 216

Zero plus one other number, 2 of each

9*3!/2! = 27

neither digit is zero, 3 of one and 1 of the other.

9C2*2*4 = 36*8 = 288

One digit is zero, 3 of one and 1 of the other.

9*3 = 27

216+27+288+27 = 558

Note: There is room here for error.

I have the same answer as Crypto, my method is just a bit different.

x=10mod 20  ....   x=10, 30, 50, ............ , 230 ......

x=5mod45     ....   x= 5, 50, 95, 140, 185, 230 ......    

since these are both true and 230-50 = 180

x= 50+180k    where k is an integer great or equal to 0

\(\frac{x}{6}=8+30k+\frac{2}{6}\\ so\\ x\mod6 = 2\)

I'd do it like this

\(4x^2 + 14x + c\\ =4(x^2+\frac{14}{4}x+\frac{c}{4})\\ =4(x^2+\frac{7}{2}x+\frac{c}{4})\\\)

now if this is a perfect square then:

\(\frac{c}{4}=[\frac{7}{2}\color{red}\div2\color{black}]\color{red}^2\\ \frac{c}{4}=[\frac{7}{4}]\color{red}^2\\ \frac{c}{4}=\frac{49}{16}\\ c=\frac{49}{4}=12\frac{1}{4}\)

LaTex:

\frac{c}{4}=[\frac{7}{2}\color{red}\div2\color{black}]\color{red}^2\\
\frac{c}{4}=[\frac{7}{4}]\color{red}^2\\
\frac{c}{4}=\frac{49}{16}\\
c=\frac{49}{4}=12\frac{1}{4}

I split it into 2 possibilities and added the probs at the end.

4hearts,                heart,    a two

\(\frac{1}{52}*\frac{12}{51}*\frac{4}{50}\)

or

4 not of hearts,    heart,     a two.

You can do this one yourself.     

Find all complex numbers z such that z^4 = 4.

Since it is a power of 4, there will be 4 solutions.

one of those solutions will be  \(\pm \sqrt[4]{ 4}=\pm \sqrt{2}\)

the other 2 will be   \(\pm \sqrt[4]{ 4}i=\pm \sqrt{2}\;i\)

Guest is certainly correct.

Another answer is 11_{base 26}

So

The smallest base 2 five digit palindrone is   10001₂     =   17₁₀

_{\(10001_2=17_{10}=33_8=11_{26}\)}

Let the midpoint of AB be M

Let angle COB = theta

Now look at right-angled triangle OMB.  

Use triangle MBO to fin an expression for cos(theta)

Use triangle BOC to find an expression for sin(theta)   (Use cosine rule)

No use sine^2(theta) + cos^2(theta) = 1

You should be able to finish it from there.   

Note that the answer is an approximation.  Maybe someone will come up with an exact answer.

That is excellent!