BWStar

I think you have the porblem backwards? We draw the red card first, so that may affect the probabilty of drawing a king after. How do I acount for this?

x = 7, y = -4

Tip:

If you don't want to solve the system of equations, simply graph the equations on something like desmos and find the intersection point. The x and  y are your answers

a+b=8

b+c=3

subtract the top by the bottom

a-c=5

a-c=5

a+c=5

add top and bottom

2a=10

a=5

now that we know a, we can find b by using the first equation

a+b=8

5+b=8

b=3

we then plug b into the second equation to get

b+c=3

3+c=3

c=0

5*3*0 = 0

abc=0

c=0

Im not too good at probability, but there is a 3/6 chance of it being a prime number (2,3,5) so that would be (3/6)^3 chance of the first 3 being prime, and then a (3/6)^3 for it not to be a prime number. so that would be (3/6)^6

According to this, your answer would be a,b = -1,3

ay = [-5,-4,-3,-2,-1]
be = [1,2,3]
lo = float('inf')
answr = 0
for a in ay:
    for b in be:
        if (1/a+1/b)*(1/b-1/a) < lo:
            lo = (1/a+1/b)*(1/b-1/a)
            answr = [a,b]
print (lo)
print (answr)

Oops, my question was what would that be equal to, in terms of n. I believe I solved with using Gauss' method and i got the sum of that equals n/4+1/(2(n+1))

Im practicing Indction, and i just want to know what the sum of that is so i can start my induction process

We replace z^4 with u^2, where u = z^2

u = z^2

y^2 = z^4

u^2 = -4

Then we take the square root of both sides

u = +-sqrt(-4)

u= i*sqrt(4) or u = -i*sqrt(4)

u = 2i or u = -2i

then we can replace u again back with z^2

z^2 = 2i or z^2 = -2i

and then we find square root again

z = 1 + i or z =  -1 - i or z = -1 + i or z = 1 + i

a+b+c =6.

There are these ways to make 6:

3+2+1

2+2+2,

We can then sort this list into 3 dfferent, 2 different, and no different.

Lets start easy.

3 same gets us only one way to permutate the numbers.         1

2 same gives us 2! ways to permutate the numbers(But we don't have any 2s)     0

and three different gives us 3! ways to permute          6

We have 7 ways to make 6 out of non zero positive whole numbers.

The answer is angle HLM bisects. I'm too lazy to do the work but if you want an explanation, find the angle measurement of MLI