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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 find all values of \( x \) for which \( f(x) > g(x) \), where \( f(x) = |x| \) and \( g(x) = x^2 \), we need to compare their values for different ranges of \( x \).

1. For \( x \geq 0 \):

\[ f(x) = |x| = x \]

\[ g(x) = x^2 \]

So, \( f(x) > g(x) \) for \( x \geq 0 \) when \( x^2 > x \), which is true when \( x > 1 \) or \( x < 0 \).

2. For \( x < 0 \):

\[ f(x) = |x| = -x \]

\[ g(x) = x^2 \]

So, \( f(x) > g(x) \) for \( x < 0 \) when \( -x > x^2 \), which is true when \( -x > x^2 \) for \( x < 0 \).

Let's solve \( -x > x^2 \) for \( x < 0 \):

\[ -x > x^2 \]

\[ x^2 + x < 0 \]

\[ x(x + 1) < 0 \]

This inequality holds true when either \( x < 0 \) and \( x + 1 > 0 \), or \( x > 0 \) and \( x + 1 < 0 \).

1. For \( x < 0 \), \( x(x + 1) < 0 \) when \( x < 0 \) and \( x + 1 > 0 \) (the product of two negative numbers is positive):

\[ x < 0 \]

\[ x + 1 > 0 \]

\[ x > -1 \]

So, for \( x < 0 \), \( f(x) > g(x) \) when \( -1 < x < 0 \).

Combining both ranges of \( x \), we have \( f(x) > g(x) \) for \( x > 1 \) or \( -1 < x < 0 \).

Therefore, all values of \( x \) for which \( f(x) > g(x) \) are \( x \in (-1, 0) \cup (1, \infty) \).

Here's how to find the number Laverne says that sends Shirley running:

Shirley's Stopping Point:

We know Shirley runs away when the sum of the numbers Laverne says exceeds 1000.

Arithmetic Series:

The sequence of numbers Laverne says forms an arithmetic series with a common difference of 5 (since she counts by 5s). The first term is 2 (as she starts with 2).

Let n be the term number:

We want to find the value of n (the term number) at which the sum of the series becomes greater than 1000.

Formula for Arithmetic Series Sum:

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

S = n/2 * (a1 + an)

where:

n = number of terms

a1 = first term

an = nth term (last term in this case)

Setting Up the Inequality:

We want the sum (S) to be greater than 1000. So, we can set up an inequality:

n/2 * (a1 + an) > 1000

Substituting Known Values:

a1 = 2 (first term)

an = a1 + (n-1) * d (nth term) where d is the common difference (5 in this case)

an = 2 + (n-1) * 5

Substitute these values into the inequality:

n/2 * (2 + (n-1) * 5) > 1000

Simplifying and Solving:

Expand the bracket:

n/2 * (2 + 5n - 5) > 1000

n/2 * (5n - 3) > 1000

Multiply both sides by 2 to eliminate the fraction:

n * (5n - 3) > 2000

Expand the left side:

5n² - 3n > 2000

Rearrange the inequality to isolate n:

5n² - 3n - 2000 > 0

Factoring the Expression:

This expression can be factored as:

(5n + 40) * (n - 50) > 0

Since the product of two factors is greater than zero only when both factors are positive or both are negative, we can consider two cases:

Case 1: (5n + 40) > 0 and (n - 50) > 0

Case 2: (5n + 40) < 0 and (n - 50) < 0

However, n (the term number) cannot be negative. So, we can disregard Case 2.

Solving Case 1:

From Case 1:

5n + 40 > 0

Subtract 40 from both sides: 5n > -40

Divide both sides by 5: n > -8

n - 50 > 0

Add 50 to both sides: n > 50

Since n represents the term number (positive integer), we are only interested in the positive solution that satisfies both inequalities.

The smallest positive integer value of n that fulfills both conditions is n = 51.

Finding the Laverne's Number:

Now that we know n (term number) is 51, we can find the corresponding number Laverne says that sends Shirley running.

Recall that the nth term (an) is calculated as:

an = a1 + (n-1) * d

an (51st term) = 2 + (51 - 1) * 5

an = 2 + 50 * 5

an = 252

Therefore, Laverne says 252 which makes the sum exceed 1000, sending Shirley running.

I got it!

Let x represent the common difference.

\(a_1+a_3+a_5+a_7+a_9+a_{11} = \frac{6(a_1+a_1+10x)}{2} = 6a_1+30x = -2 \)

\(a_2+a_4+a_6+a_8+a_{10}+a_{12} = \frac{6(a_1+x+a_1+11x)}{2} = 6a_1+36x = 1\)

Solving the equation gives us \(x = \frac{1}{2}, a_1 = -\frac{17}{6}\)

From the given information, we know that there are 3 students in the math club, and there are twice as many students in the science club as in the math club which means there are \(3 \times 2 = 6\) students in the science club. However, we cannot simply add 6 to 3 because there are students who are in both clubs. There are 3, in this case, based on the question. Therefore, we need to subtract 3 from \(6 + 3 \) which will be 6.

Let's denote the common difference of the arithmetic sequence by \(d\).

For the sum of an arithmetic series, we have the formula:

\[ S = \frac{n}{2} \cdot (a_1 + a_n) \]

Given that \(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2\), we can write the sum as:

\[ S_1 = \frac{6}{2} \cdot (a_1 + a_{11}) = 3 \cdot (a_1 + (a_1 + 10d)) = 3 \cdot (2a_1 + 10d) \]

\[ S_1 = 6a_1 + 30d \]

Similarly, for \(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1\), we get:

\[ S_2 = \frac{6}{2} \cdot (a_2 + a_{12}) = 3 \cdot (a_2 + (a_2 + 11d)) = 3 \cdot (2a_2 + 11d) \]

\[ S_2 = 6a_2 + 33d \]

Given that \(a_1 + a_3 + a_5 + a_7 + a_9 + a_{11} = -2\) and \(a_2 + a_4 + a_6 + a_8 + a_{10} + a_{12} = 1\), we can write the following system of equations:

\[ 6a_1 + 30d = -2 \]

\[ 6a_2 + 33d = 1 \]

Now, let's solve this system of equations. Subtracting the first equation from the second, we get:

\[ 6a_2 + 33d - (6a_1 + 30d) = 1 - (-2) \]

\[ 6a_2 - 6a_1 + 3d = 3 \]

\[ 6(a_2 - a_1) + 3d = 3 \]

\[ 6d + 3d = 3 \]

\[ 9d = 3 \]

\[ d = \frac{1}{3} \]

Now, substitute \(d = \frac{1}{3}\) into one of the original equations to find \(a_1\).

Let's use the first equation:

\[ 6a_1 + 30 \left(\frac{1}{3}\right) = -2 \]

\[ 6a_1 + 10 = -2 \]

\[ 6a_1 = -12 \]

\[ a_1 = \boxed{-2} \]

So, \(a_1 = -2\).

12 choose 5 is 792.

Here's how to find the number of times the digit 9 appears in the list of all integers from 1 to 750:

Approach 1: Separating by Hundreds Place

We can break down the count into three sections based on the hundreds digit:

Numbers from 1 to 99:

Every unit digit (from 0 to 9) will appear 10 times (once in each number from 10 to 19, 20 to 29, and so on).

Since 9 is a unit digit, it appears 10 times in this range.

Numbers from 100 to 199:

The hundreds digit is always 1, so we only need to consider the tens and units digits.

Following the same logic as before, each digit (including 9) will appear 10 times in the tens place.

Therefore, 9 appears 10 times in this range.

Numbers from 200 to 750:

We again need to consider both the hundreds and tens/units digits.

The hundreds digit can be 2, 3, ..., 7 (6 different values).

For each hundreds digit value, the tens and units digits will each allow the digit 9 to appear 10 times (as explained earlier).

Counting the Occurrences:

Occurrences in 1-99: 10 times

Occurrences in 100-199: 10 times

Occurrences in 200-750: 6 hundreds digits * 10 occurrences/hundreds digit = 60 times

Total Occurrences:

Adding the occurrences from each section:

Total = 10 (from 1-99) + 10 (from 100-199) + 60 (from 200-750) = 80 times

Approach 2: Utilizing Digit Patterns

Another approach is to analyze the digit patterns within the range.

Single-digit numbers (1-9): The digit 9 appears once.

Two-digit numbers ending in 9 (from 19 to 99): Each number contributes the digit 9 once. There are 10 such numbers (9 total for 10s digit and 1 for the unit digit of 99).

Three-digit numbers with 9 in the tens or units place (from 109 to 759, and from 190 to 799): Similar to two-digit numbers, each number contributes the digit 9 once. There are 10 numbers for each placement of 9 (tens or units), resulting in 20 numbers.

Three-digit numbers with 9 in the hundreds place (from 900 to 999): Each number contributes the digit 9 twice. There are 100 such numbers (10 numbers from 900 to 909 and another 90 numbers from 910 to 999).

Counting the Occurrences:

Single digit: 1 time

Two-digit ending in 9: 10 times

Three-digit with 9 in tens/units: 20 times

Three-digit with 9 in hundreds: 100 times (each digit counted twice)

Total Occurrences:

Adding the occurrences from each pattern:

Total = 1 + 10 + 20 + 2 * 100 = 131 times

Addressing the Discrepancy:

The second approach seems to give a higher count (131) compared to the first approach (80). However, there's a double-counting error in the second approach.

Numbers like 199 and 909 are counted twice (once for 9 in tens place and again for 9 in units place).

Correcting the Count:

We need to subtract the number of times a three-digit number is counted twice for both tens and units digits containing 9.

There are 20 such numbers (as counted earlier).

Final Count:

Total occurrences (corrected) = 131 (initial count) - 20 (double-counted) = 111 times

This corrected count (111) is incorrect. The first approach (80) provides the accurate answer.

The mistake in the second approach highlights the importance of careful analysis and avoiding double-counting when dealing with digit patterns.

Here's how to find the number of blue marbles in the bag:

Total Ratio:

We are given that the ratio of yellow, green, blue, and white marbles is 4:1:2:3. This represents the proportion of each color relative to the total number of marbles.

The sum of these ratios represents the total number of parts that make up the whole bag. In this case:

Total Ratio = 4 + 1 + 2 + 3 = 10

Fraction per Marble Color:

To find the fraction representing the proportion of each color, we can divide each value in the ratio by the total ratio:

Fraction of Yellow = 4 / 10 = 2/5

Fraction of Green = 1 / 10

Fraction of Blue = 2 / 10 (what we're solving for)

Fraction of White = 3 / 10

Total Number of Marbles:

The problem states that the bag contains 20 marbles.

Number of Blue Marbles:

Since the fraction of blue marbles (2/10) represents their proportion of the total number of marbles (20), we can set up a proportion to find the actual number of blue marbles (b):

Fraction of Blue : Total Number of Marbles = Number of Blue Marbles : Total Number of Marbles

Substituting the known values:

(2/10) : 20 = b : 20

Solving for b (Number of Blue Marbles):

Cross-multiplying:

2 * 20 = 10 * b

Divide both sides by 10:

b = 4

Therefore, there are 4 blue marbles in the bag.

Properties of the Rhombus:

Since the given rhombus has all sides equal (4 cm) and two acute angles of 90 degrees each, it's essentially a square.

Diagonals of the Square:

The diagonals of a square bisect each other at right angles and divide the square into four congruent right triangles. Each diagonal has a length equal to the square root of twice the side length squared (sqrt(2s^2)).

In this case, the side length (s) is 4 cm. Therefore, the length of each diagonal (d) is:

d = sqrt(2 * s^2)

d = sqrt(2 * 4^2 cm^2)

d = sqrt(32) cm

d = 4√2 cm (since the square root of 32 can be simplified to 4 times the square root of 2)

Radius of the Inscribed Circle:

The inscribed circle in a square touches the midpoint of each side. Since the diagonals bisect each other at the center of the square and are perpendicular, they also pass through the center of the inscribed circle.

Therefore, the radius (r) of the inscribed circle is half the length of a diagonal divided by two (since there are two diagonals and the circle touches them both at their midpoints):

r = (d / 2) / 2

r = (4√2 cm) / 2 / 2

r = √2 cm

Answer:

The radius of the inscribed circle is √2 centimeters.

To compute \( \log_{8^a} 4^b \) in terms of \( a \) and \( b \), we'll utilize the properties of logarithms.

First, let's express \( 8^a \) and \( 4^b \) in terms of a common base. Since both 8 and 4 can be represented as powers of 2, we rewrite them as follows:

\[ 8^a = (2^3)^a = 2^{3a} \]

\[ 4^b = (2^2)^b = 2^{2b} \]

Now, our expression becomes:

\[ \log_{8^a} 4^b = \log_{2^{3a}} 2^{2b} \]

Using the property of logarithms that \( \log_a b^n = n \cdot \log_a b \), we rewrite this as:

\[ = 2b \cdot \log_{2} 2^{3a} \]

Since \( \log_{2} 2^{3a} = 3a \) (as \( \log_{a} a = 1 \)), our expression simplifies to:

\[ = 2b \cdot 3a \]

Thus, \( \log_{8^a} 4^b \) in terms of \( a \) and \( b \) is \( 6ab \).