ChiIIBill

To solve the inequality \( |x - 4| + 2|x + 3| \leq 11 \), we'll consider different cases depending on the sign of \( x - 4 \) and \( x + 3 \).

Case 1: \( x - 4 \geq 0 \) and \( x + 3 \geq 0 \)

In this case, both absolute values are positive, so we have:

\[ (x - 4) + 2(x + 3) \leq 11 \]

\[ x - 4 + 2x + 6 \leq 11 \]

\[ 3x + 2 \leq 11 \]

\[ 3x \leq 9 \]

\[ x \leq 3 \]

Case 2: \( x - 4 \geq 0 \) and \( x + 3 < 0 \)

In this case, \( x - 4 \geq 0 \) and \( x + 3 < 0 \), so \( x \) is between -3 and 4.

\[ (x - 4) + 2(-x - 3) \leq 11 \]

\[ x - 4 - 2x - 6 \leq 11 \]

\[ -x - 10 \leq 11 \]

\[ -x \leq 21 \]

\[ x \geq -21 \]

Case 3: \( x - 4 < 0 \) and \( x + 3 \geq 0 \)

In this case, \( x - 4 < 0 \) and \( x + 3 \geq 0 \), so \( x \) is between -3 and 4.

\[ -(x - 4) + 2(x + 3) \leq 11 \]

\[ -x + 4 + 2x + 6 \leq 11 \]

\[ x + 10 \leq 11 \]

\[ x \leq 1 \]

Case 4: \( x - 4 < 0 \) and \( x + 3 < 0 \)

In this case, both absolute values are negative, so we have:

\[ -(x - 4) + 2(-x - 3) \leq 11 \]

\[ -x + 4 - 2x - 6 \leq 11 \]

\[ -3x - 2 \leq 11 \]

\[ -3x \leq 13 \]

\[ x \geq -\frac{13}{3} \]

Now, let's combine the solutions from each case:

- Case 1: \( x \leq 3 \)

- Case 2: \( x \geq -21 \)

- Case 3: \( x \leq 1 \)

- Case 4: \( x \geq -\frac{13}{3} \)

The overlapping intervals are \( [-21, 3] \) and \( \left[ -\frac{13}{3}, 1 \right] \).

Therefore, the values of \( x \) satisfying the inequality are \( \boxed{[-21, 3] \cup \left[ -\frac{13}{3}, 1 \right]} \).

To find the possible values of the tens digit of a perfect square with a units digit of 6, we need to consider the squares of integers ending in 4 or 6, as these are the only possibilities for a number to have a units digit of 6 when squared.

Let's denote the perfect square as \( n^2 \), where \( n \) is an integer.

1. If the units digit of \( n \) is 4, then the units digit of \( n^2 \) will be 6.

   - For example, if \( n = 24 \), then \( n^2 = 576 \).

2. If the units digit of \( n \) is 6, then the units digit of \( n^2 \) will also be 6.

   - For example, if \( n = 26 \), then \( n^2 = 676 \).

From these examples, we can observe that the tens digit of \( n^2 \) can be either 2 or 6.

Therefore, the possible values of the tens digit of a perfect square with a units digit of 6 are \( \boxed{2, 6} \).

Using an interest payment calculator, each payment is 32.51, so the total cost is 18*32.51 = 585.18.

We can solve for S_15, the sum of the first 15 terms, using the properties of arithmetic series and the given information about S_5 and S_10.

Here's how to approach this problem:

Formula for Arithmetic Series Sum:

The sum (S_n) of an arithmetic series can be calculated using the formula:

S_n = n/2 * (a_1 + a_n)

where:

n = number of terms

a_1 = first term

a_n = nth term (last term in this case)

Relating S_5 and S_10:

We are given that:

S_5 = 1/5 (sum of the first 5 terms)

S_10 = 1/10 (sum of the first 10 terms)

Finding the Difference (S_10 - S_5):

Subtracting S_5 from S_10, we can eliminate the first term (a_1) from the equation:

S_10 - S_5 = (1/10) - (1/5)

This represents the sum of terms from the 6th term (a_6) to the 10th term (a_10) of the arithmetic sequence.

Simplifying the Difference:

(1/10) - (1/5) = (1 - 2) / 10 = -1/10

Understanding the Difference:

The difference (-1/10) represents the sum of the next 5 terms (a_6 to a_10) after the first 5 terms (a_1 to a_5). Since it's negative, it implies that the common difference (d) between terms in the sequence is negative.

Finding the Sum of Next 5 Terms (S_10 - S_5):

We can express the difference (-1/10) using the formula for the sum of an arithmetic series with n = 5 (number of terms from 6th to 10th):

-1/10 = 5/2 * (a_6 + a_{10})

Since we know the common difference (d) is negative, the sum of the next 5 terms (a_6 + a_{10}) is also negative.

Solving for S_15:

To find S_15 (sum of the first 15 terms), we can build upon the relationship between S_10 and S_5:

S_15 = S_10 + (Sum of terms from 11th to 15th)

We already know S_10 (1/10) and the relationship between S_10 and S_5 (difference representing the sum of terms from 6th to 10th).

The sum of terms from the 11th to 15th term will be similar to the sum of terms from the 6th to 10th term (both sets of 5 terms with a negative common difference).

Therefore, the sum of terms from the 11th to 15th term will also be -1/10.

Finding S_15:

S_15 = (1/10) + (-1/10) = 0

Therefore, the sum of the first 15 terms (S_15) is 0.

To avoid ever having the same group of 5 members review a book, we need to ensure there are enough total members (z) such that any group of 5 can be chosen without repeating a combination.

Here's the key idea:

We care about the number of distinct groups of 5 members we can form, not just the total number of possible selections.

Combinations vs. Permutations:

Combinations: Order doesn't matter (e.g., John, Mary, Sarah is the same group as Sarah, John, Mary).

Permutations: Order matters (e.g., John reviewing first is different from Mary reviewing first).

In this case, since the order the members review the book doesn't matter, we're interested in the number of combinations of 5 members we can choose from a group of z people.

Using Combinations Formula:

The number of ways to choose 5 members out of z for a book review can be calculated using the combinations formula:

nCr = n! / (r! * (n-r)!)

where:

n = Total number of members (z)

r = Number of members chosen for review (5 in this case)

Finding the Smallest z:

We want to find the smallest positive integer z such that the number of combinations of choosing 5 members (nCr) is greater than or equal to the total number of days (400). In other words:

nCr >= 400

Trial and Error with Combinations:

We can try different values of z and calculate the corresponding nCr using the formula. The smallest z that satisfies the condition will be our answer.

For z = 5 (5 choose 5): 1 (There's only one group possible - all 5 members) - not enough.

For z = 6 (6 choose 5): 6 (We can choose 5 members out of 6 in 6 ways) - still not enough.

For z = 7 (7 choose 5): 21 (We can choose 5 members out of 7 in 21 ways) - finally enough!

Therefore, the smallest positive integer z is 7.

The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.

x + y + 3z  =  10     --->     x -2     --->     -2x - 2y - 6z  =  -20

4x + 2y + 5z  =  7       --->                      4x + 2y + 5z  =  7

Add down the columns:                             -6x - z  =  -13   

Since  kx + z  =  3        --->         z  =  3 - kx

Substituting these:        -6x - (3 - kx)  =  -13

                                    -6x - 3 + kx  =  -13

                                      6x + 3 - kx  =  13

                                          6x - kx  =  10

                                          x(6 - k)  =  10

If  6 - k  =  0,  there is no value for x that will result in a product of 10, so  k = 6  results in no possible solution.