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(a) The sum of the numbers in the first row is 1/(1 - a).

The sum of the numbers in the second row is ab/(1 - a).

The sum of the numbers in the third row is (a^2 b^2)/(1 - a).

So, the sum of the numbers in the rows form a geometric sequence, which adds up to

1/(1 - a) + ab/(1 - a) + (a^2 b^2)/(1 - a) + ...  = 1/((1 - a)(1 - ab)).

(b) Since the colors of the chessboard alternate white and black, the sum of the numbers on the black squares is equal to the sum of the numbers on the white squares, except for the numbers that are on every other white square, and every other black square.

The sum of the numbers that are on every other white square is a/((1 - a)(1 - ab)), and the sum of the numbers that are on every other black square is b/((1 - a)(1 - ab)), so to find the sum of the numbers on the black squares, we take half the difference, which gives us a sum of

1/((1 - a)(1 - ab)) + 1/2*a/((1 - a)(1 - ab)) + 1/2*b((1 - a)(1 - ab)) = (a + b + 2)/(2(1 - a)(1 - ab)).

it will converge to 1.25

no problem 

My counting and probability is not very good but I'll try. I think this is how to do the problem:

We look at where the teacher chooses all girls in the group. It's a group of 3 and there are 4 girls so there are 4C3=4 ways to choose 3 girls(a group). Then we look at the total number of groups the teacher can choose which is 10C3=120. Therefore the probability everyone in the group is a girl is 4/120=1/30.

Beats me ....   What did YOU find after I told you HOW to get the answer?   I'm not here to puke up answers to people's homework

If the hypotenuse (XY) is 25 then the legs (XY and ZY) must be  20 and 15, and the height (ZW) is 12

So what's the answer?

I'm still stuck. Can you explain farther?

Nice figure

The midsegment is the line segmnet MN where M is the midpoint of (1,1) and (2,8) and N is the midpoint of (5,1) and (4,8).

M = (1.5, 4.5)  and  N = (4.5, 4.5).

Now, use the slope formula.

Using the graphing equation:  y  =  a(x - h)² + k     where  (h, k) is the vertex.

The vertex of this parabola occurs at  (2, 1)     --->     y  =  a(x - 2)² + 1

It rises, so that the value of a is positive.

It has no stretching or contracting so the value of a is 1.

The equation is:  y  =  1(x - 2)² + 1     or     y  =  (x - 2)² + 1.

Now, multiply this out and simplify ... 

Arrange first one as    x = y^2 -8y+13           then the 'y' value of the vertex will be found   (this is a sideways parabola)

                                                                  as   - b/2a     where b = - 8  a =1     use THIS value in the equation to find the 'x' value

y = -2x^2 + 8x -15        vertex 'x' value will  be   -b/2a   b = 8     a = -2      use THIS value in the equation to find the vertex 'y' value

A + B + C  =  900

C  =  2A          B  =  A + 100

A + (A + 100) + (2A)  =  900

                  4A + 100  =  900

                            4A  =  800

                              A  =  200

B  =  A + 100  =  200 + 100  =  300

g(x)  =  15x² + 12x - 13

g(4)  =  15·4² + 12·4 - 13  =  275

f(x)  =  11x - 15

f( g(4) )  =  f( 275 )  =  11·275 - 15  =  ....

log_x[ x^1/2 / x ]  =  a     --->     x^1/2 / x  =  x^a 

Since x must be positive:  x^1/2 / x  =  x^-1/2  =  x^a     --->     a  =  -1/2

                                                                                --->    log_x[ x^1/2 / x ]  =  -1/2 

The answer is 316,251.

We can start with casework to get an idea:
Case one: Everyone (all fifty) vote: this is 53 choose 3. This is because we are employing the stars and bars method where the "stars" are the votes total, and the bars divide them up into 4 chunks for 4 candidates.

Case two: One person decides not to vote: this is 52 choose 3.

We can continue like this, and when we add the cases up we will end up with something like this:

\({53\choose3}+{52\choose3}+{51\choose3}+\cdots+{4\choose3}.\)

(that last case is when only one person votes, and we have 1 star and 3 bars and need to find 3 places to put 3 bars).

This looks familiar!

If we re-arrange it:

\({4\choose1}+\cdots+{51\choose49}+{52\choose 48}+{53\choose47}\)

It's the hockey stick identity!

All we're missing is the 3 choose 0 term. This can be solved easily:

\({4\choose1}+\cdots+{51\choose49}+{52\choose48}+{53\choose47}={54\choose47}-{3\choose0}={54\choose3}-1\)

(2-sqrt5)^2 = 4 - 4 sqrt5 +5    now multiply it again

(9 - 4 sqrt 5)(2-sqrt5) =    18 -17sqrt 5 +20 =    38 - 17 sqrt5    

I'll start with y - mx + b form      m = slope =   (y1-y2)/(x1-x2)    =  -4

y = -4x + b     sub in one of the points to calc 'b'

-5 = -4(2) + b     b = 3

y = -4x+3    now convert to standard form

y + 4x = 3

You can just 'brute' force it     divide by 6   four coins left over

                 10     16     22     28     34     Divide these by 5 to see leftover (looking for 3)

leftover       0       1      2       3 

So 28 coins    divide by 7    ZERO leftovers

The pole is 9 meters

Sorry, that's incorrect. Can you expand more behind your reasoning so I can attempt your method myself?