learnmgcat

There are 650 ways.

The answer is c = 18/11.

The probability is 3/16.

The answers are 47 and 22.

The answer is r  = n^2/(n^2 - 1).

The number of paths is 140.

We can rewrite the expression to make simplification easier:

M = 9^40000 / (9^200 - 2) = (9^200)^200 / (9^200 - 2)

Since 9^200 is a very large number, when we divide by (9^200 - 2), the remainder will be very close to the remainder of dividing 1 by (9^200 - 2).

Now, let's analyze (9^200 - 2):

The units digit of 9^n repeats in a cycle of 1, 9, 8, 7, 6, 5, 4, 3, 2 (powers of 9 ending in 1, 9, 8, etc.). Since 200 is a multiple of 4, the units digit of 9^200 will be 1.

Subtracting 2 from 1 results in a units digit of -1, which is equivalent to 9 (since remainders are non-negative).

Therefore, dividing 1 by (9^200 - 2) will have a remainder of 1.

When we take M, a very large number divided by (9^200 - 2) with a remainder of 1, rounding it to the nearest integer won't change the remainder when divided by 100.

So, the remainder of M divided by 100 will also be 9​.

For problem 1, we notice that both equations have the same coefficient on the term 2b.

This means that in order for the system to have infinitely many solutions, the coefficients of a must also be proportional.

We can rewrite the second equation as: 3a+2b=k+mb. Thus, for infinitely many solutions, we need [ \frac{4}{3} = \frac{-1}{m}. ]

Since the fraction on the left-hand side is positive, we know that m must be negative. Solving for m, we get m = −3​. Then k = 4/3​⋅−3 + 2​ = −2.

Therefore, k = -2 and m = -3.