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To solve this problem, let's focus on the ways we can get x in the problem. 

First, we have x and 14. 

We get \(14x\) from this. 

Next, we have 1 and \(-23x\)

Multiplying these two together, we get

\(-23x\)

Adding these two together, we get \(14x-23x=-9x\)

We could also expand everything out. Using the Distributive Property, we get

\(x^{7}-7x^{6}+10x^{5}-13x^{4}+8x^{2}-9x+14\)

Thus, the coefficient is -9. 

So our answer is -9. 

Thanks! :)

In this case, note that the only way x is possibily not in the domain is if the denominator of a fraction is 0. 

First off, the smallest fraction 3/x. x can't be 0 since 3/0 is undefined. 

Next, we have

\(1+3/x \neq 0\\ 3/x\neq-1\\ -x \neq 3\\ x \neq -3\)

Thus, x cannot be negative 3. Lastly, let's focus on the last fraction. We have

\(1+\frac{2}{1+3/x} \neq 0\\ \frac{2}{1+3/x}\neq-1\\ 2\neq-1-3/x\\ 3\neq -3/x\\ x \neq -1\)

So x also can't possibly be -1. 

Thus, we have 3 following solutions. We have x cannot be 0, -1 and -3. 

So our answer is 0, -1, -3. 

Thanks! :)

The only way for the system to have infinitely solutions is where we have two exactly congruent equations. 

We first notice that we have

\(6a + 2b = k + 3a + mb\)

Subtracting 3a from both sides will achieve us the same exact left sides of the equations. We have

\(3a+2b=2\\ 3a+2b = k+mb\)

Setting the right sides of the equation to equal each other, we have

\(k+mb=2\)

Now, let's note that if m was any number other than 0, the two equations would fail to match each other. Therefore, m must be 0. 

For this value of m, we have k is 2. So we have

\(m=0\\ k=2\)

So our final answer is m = 0 and k = 2. 

Thanks! :)

First, let's put y in terms of x to make our lives easier. 

We have

\(y=25-x\)

Now, we subsitute it into the second expression and see what we get, 

We have

\(6x+3(25-x)\\ 6x+75-3x\\ 3x+75\)

Now, since x must be nonegative, the smallest value it can be is 0. Thus, we have

\(3(0) + 75\\ 75\)

So 75 is our answer. 

Thanks! :)

The probability is 1/8.

The probability is 5/12.

The expected value is 5/4.

The answer is 3*sqrt(2).

There are many ways to solve this problem, but I prefer casework. 

There are two cases. 

The first case is that we get two even numbers from the lists. We have

\(\frac{\binom22}{\binom52} = 1/10\)

The second case we have is that we one odd number and we have one even number. Using permutations, we get

\(\frac{\binom31}{\binom52} = 6/10\)

Adding these two together, we get

\(1/10+6/10=7/10\)

So our final answer is 0.7

Thanks! :)

First, we can use variables to solve this problem in  quite a handy way. First, let's set

\(x = 2.\overline{57}\)

For this value of x, we know that

\(100 x= 257.\overline{57}\)

Now, we do the really cool trick. We subtract the first equation from the second equation to get

\(100 x - x = 257.\overline{57} - 0.\overline{57}\)

Notice that the repeating terms cancel out. Thus, we have

\(99x =257\\ x= \frac{255}{99}\)

Dividing by a common factor, we have

\(x= 2 \frac{19}{33}\)

Thus, our final answer is \(2.\overline{57} = 2 \frac{19}{33}\)

Thanks! :)

What is the largest positive integer $n$ such that $1457$, $1797$, $709$, $15$, $24$, $197$, $428$ all leave the same remainder when divided by $n$?   

First, let's eliminate as many divisors as we can.  Consider 15 and 24.   

The divisor n has to leave a remainder, so factors can be eliminated.   

n cannot be 3 or 5, nor 2, 4, 6, 8, or 12 .... leaving 7, 9, 10, 11, 13, and 14.   

Remember, the remainder has to be the same.  Try them out on 15 and 24 first.  

15/7 = 2 R 1      24/7 = 3 R 3      so it isn't 7     

15/9 = 1 R 6      24/9 = 2 R 6      so 9 looks hopeful, we'll try another   

1457/9 = 161 R 8                        so 9 is eliminated   

15/10 = 1 R 5     24/10 = 2 R 4    so it isn't 10   

15/11 = 1 R 4     24/11 = 2 R 2     so it isn't 11   

15/13 = 1 R 2     24/13 = 1 R 11   so it isn't 13   

15/14 = 1 R 1     24/14 = 1 R 10   so it isn't 14   

Well, I have run out of divisors and ideas at the same time.   

Either the problem has no solution, or I'm missing something.   

_.

Let's consider the card labeled as follows:

Aces: \(A_1,A_2,A_3,A_4\)

Non-Aces: \(B_1,B_2 \)

The total number of ways to choose 2 cards out of 6 for the first pile is \(\binom{6}{2} \). After choosing the first pile, we choose 2 cards out of the remaining 4 for the second pile, which is \(\binom{4}{2} \). The remaining 2 cards automatically form the third pile, and this can be done in \(\binom{2}{2} \) ways.

Total ways is: \(\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}\)

But we have to divide 3!, since the order of the piles doesn't matter.

So we get \(\frac{\binom{6}{2}\times\binom{4}{2}\times\binom{2}{2}}{6}\) = 15

Next, we count the number of favorable outcomes where each pile contains exactly one Ace. Each pile must have one Ace and one non-Ace. There are 4 Aces (\(A_1,A_2,A_3,A_4\)) and 2 Non-Aces (\(B_1,B_2 \)).

The assignment of Aces to each pile can be done in 4 ways for first pile 3 ways for second 2 for final simplifying: \(4\times3\times 2\) = 24 ways

Every non-ace gets included ensuring simplified piles count as above:

\(\boxed{\frac{4}{15}}\)

Apply Burnside's Lemma: Number of distinct colorings = 1/4*(∣C_0 degrees​∣ + ∣C_90 degrees​∣ + ∣C_180 degrees​∣ + ∣C_270 degrees​∣)

For C_0 degrees: No squares are moved so it is 2^16

For C_90 degrees: Since there are 4 quadrants, it is 2^4

For C_180 degrees: Each square must match the one diametrically opposite to it. This creates 8 pairs of squares, where each pair must be the same color, so it is 2^8

For C_270 degrees: You do the same thing as C_90 degrees: 2^4

So the answer is 1/4*(65536 + 16 + 256 + 16) = 16448

(I'm not fully sure about this answer, but if it is wrong please tell me) :)

The probability is 9/13.

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.