blackpanther

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 #2
avatar+1352 
0

To find the largest possible value of \( CF \), we'll use the relationship between the altitudes in a triangle and its area. The area \( \Delta \) of a triangle can be expressed using any of its altitudes:

 

\[
\Delta = \frac{1}{2} \times \text{base} \times \text{height}
\]

 

For triangle \( ABC \), we have the following relationships:

 

- \( \Delta = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times CA \times BE = \frac{1}{2} \times AB \times CF \)

 

Let the side lengths be \( a = BC \), \( b = CA \), and \( c = AB \). Then:

 

\[
\Delta = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times CF
\]

 

This simplifies to:

 

\[
\Delta = 6a = 8b = \frac{1}{2} c \times CF
\]

 

Equating these expressions:

 

\[
6a = 8b \quad \text{and} \quad 6a = \frac{1}{2} c \times CF
\]

 

### Step 1: Solve for \( a \) and \( b \)


From \( 6a = 8b \):

 

\[
\frac{a}{b} = \frac{8}{6} = \frac{4}{3}
\]

 

Thus, \( a = \frac{4}{3}b \).

 

 

### Step 2: Substitute into \( 6a = \frac{1}{2} c \times CF \)


Substitute \( a = \frac{4}{3}b \) into \( 6a = \frac{1}{2} c \times CF \):

 

\[
6 \times \frac{4}{3}b = \frac{1}{2} c \times CF
\]

 

Simplifying:

 

\[
8b = \frac{1}{2} c \times CF \quad \text{so} \quad 16b = c \times CF
\]

 

### Step 3: Find the Maximum Value of \( CF \)


We need to maximize \( CF \), which is a positive integer. Since \( 16b = c \times CF \), and \( c \) and \( CF \) are integers, \( CF \) is maximized when \( c \) is minimized.

 

Given the relationship \( 6a = 8b \), \( a = \frac{4}{3}b \), and \( c = \frac{16b}{CF} \), the smallest integer value of \( c \) occurs when \( CF \) is as large as possible.

 

If \( CF = 16 \), then:

 

\[
16b = c \times 16 \quad \text{so} \quad c = b
\]

 

Since \( c \) is minimized and \( CF \) is maximized at \( 16 \), this is the largest possible value for \( CF \).

 

Thus, the largest possible value of \( CF \) is \( \boxed{24} \).

14.08.2024
 #1
avatar+1352 
+2

Let's analyze the given information and set up an equation to find the desired number of palindromes.

 

Three-digit palindrome: We can represent the number as abc=100a+10b+c, where a is the hundreds digit, b is the tens digit, and c is the units digit (since it's a palindrome, a=c ).

 

Sum of digits and product: The problem states that the sum of the hundreds digit, units digit, and the product of the units and tens digits is eight more than the tens digit. Translating this into math, we get: a+c+bc=b+8.

 

Solving the equation: Since a=c, we can substitute to get: $$ a + a + ab = b + 8.$$ Combining like terms: $$ 2a + ab = b + 8.$$ Factoring out b from the left side: b(a+1)=b+8.

 

Now we can analyze the solutions for b:

 

If b=0, then the equation is always true regardless of a. However, a three-digit palindrome cannot have a leading zero (hundreds digit cannot be zero). So, b=0.

 

If b=0, then we can divide both sides by b to get: a+1=1+b8​. Since a is a digit (0-9), the expression on the right side must be an integer. This is only possible when b is a divisor of 8.

 

Possible values for b are: b=1,2,4,8. Let's check each case to see if it leads to a valid three-digit palindrome:

 

b=1: In this case, a=0 (which we cannot have).

 

b=2: Here, a=3, resulting in the palindrome 323​.

 

b=4: We get a=1, which leads to the palindrome 141​.

 

b=8: This results in a=−1, which is not a valid digit.

 

Therefore, there are only two three-digit palindromes that satisfy the given property: 323 and 141.

30.06.2024
 #1
avatar+1352 
0

Here's the solution for 1.

 

We can approach this problem by analyzing the given inequality and maximizing the expression for x^2 + 5x + 6 within the constraints.

 

Factoring the Inequality: The inequality x^2 + 7x + 12 <= 0 can be factored as (x + 3)(x + 4) <= 0. This means either both factors are less than or equal to zero, or one factor is positive and the other is negative (the product becomes zero).

 

Identifying the Range of x: From the factored inequality, we know:

 

x ≤ -3 (when x + 3 is less than or equal to zero)

 

x ≤ -4 (when x + 4 is less than or equal to zero)

 

However, the second condition (x ≤ -4) is redundant as -3 is already less than or equal to -4. Therefore, the valid range for x is x ≤ -3.

 

Maximizing x^2 + 5x + 6: We want to maximize the value of x^2 + 5x + 6 within the constraint x ≤ -3. We can rewrite the expression as a completed square:

 

x^2 + 5x + 6 = (x^2 + 5x + 25/4) + 11/4 = (x + 5/2)^2 + 11/4

 

Since the square of any real number (x + 5/2) is non-negative (including zero), adding a constant positive value (11/4) will always result in a positive or zero value.

 

Largest Possible Value: The largest possible value for x^2 + 5x + 6 occurs when (x + 5/2)^2 is zero. This happens when x = -5/2. However, this value of x (-5/2) violates the constraint x ≤ -3.

 

Therefore, the largest possible value for x^2 + 5x + 6 within the allowed range of x is achieved when x is as close to -3 as possible (without exceeding it). This occurs when x = -3.

 

Evaluating the Maximum Value: Substituting x = -3 in the expression:

 

x^2 + 5x + 6 = (-3)^2 + 5(-3) + 6 = 9 - 15 + 6 = 0

 

Answer: The largest possible value of x^2 + 5x + 6 is 0.

25.05.2024
 #2
avatar+1352 
0

There are three cases to consider, since all three marbles can be red, all white, or all blue.

 

Case 1: All Red Marbles

 

Favorable cases: We need to choose 3 red marbles from 4 available. This can be done in ⁴C₃ ways (combinations, not permutations).

 

Total cases: We can choose any 3 marbles from a total of 4 red + 5 white + 6 blue = 15 marbles in ¹⁵C₃ ways.

 

Case 2: All White Marbles

 

Favorable cases: We need to choose 3 white marbles from 5 available. This can be done in ⁵C₃ ways.

 

Total cases: Same as Case 1 (¹⁵C₃).

 

Case 3: All Blue Marbles

 

Favorable cases: We need to choose 3 blue marbles from 6 available. This can be done in ⁶C₃ ways.

 

Total cases: Same as Case 1 (¹⁵C₃).

 

Total Probability

 

The probability that all three marbles are the same color (considering all three cases) is the sum of the probabilities of each case:

 

P(all same) = P(all red) + P(all white) + P(all blue)

 

Calculate Each Probability:

 

P(all red) = ⁴C₃ / ¹⁵C₃ P(all white) = ⁵C₃ / ¹⁵C₃ P(all blue) = ⁶C₃ / ¹⁵C₃

 

We can simplify each term using the fact that nCr = n! / (r! * (n-r)!):

 

P(all red) = (4!) / ((3!) * (1!)) / ((15!) / ((3!) * (12!))) P(all white) = (5!) / ((3!) * (2!)) / ((15!) / ((3!) * (12!))) P(all blue) = (6!) / ((3!) * (3!)) / ((15!) / ((3!) * (12!)))

 

Notice that (3!) and (12!) terms appear in all the denominators and numerators (except the first factorials in the numerators). These terms will cancel out, leaving:

 

P(all red) = 4 / (15 * 14 * 13) P(all white) = 10 / (15 * 14 * 13) P(all blue) = 20 / (15 * 14 * 13)

 

Final Probability:

 

Add the probabilities of each case:

 

P(all same) = 4/210 + 10/210 + 20/210 = 34/210

 

Simplify the fraction:

 

P(all same) = 17/105

 

Therefore, the probability that all three marbles drawn are the same color is 17/105.

09.05.2024
 #2
avatar+1352 
0

To find the probability of winning a super prize in the SuperLottery, we need to consider two mutually exclusive events:

 

1. **Winning by matching at least two of the white balls.**


2. **Winning by matching the red SuperBall.**

 

We'll calculate the probabilities for each event and then add them together.

 

1. **Winning by matching at least two of the white balls:**

 

   To calculate this probability, we can find the probability of not matching any of the white balls and subtract it from 1.

 

   The probability of not matching any of the white balls on a single draw is:

 

   \[\frac{{9 \choose 3}}{{12 \choose 3}}\]

 

   So, the probability of matching at least two of the white balls is:

 

   \[1 - \frac{{9 \choose 3}}{{12 \choose 3}}\]

 

2. **Winning by matching the red SuperBall:**

 

   The probability of matching the red SuperBall is simply \( \frac{1}{8} \) since there's only one SuperBall drawn from 8 possibilities.

 

Now, let's calculate these probabilities:

 

1. Probability of winning by matching at least two of the white balls:


   \[1 - \frac{{9 \choose 3}}{{12 \choose 3}} = 1 - \frac{84}{220} = 1 - \frac{21}{55} = \frac{34}{55}\]

 

2. Probability of winning by matching the red SuperBall:


   \[P(\text{Red SuperBall}) = \frac{1}{8}\]

 

Finally, to find the total probability of winning a super prize, we add the probabilities of the two mutually exclusive events:

 

\[P(\text{Winning super prize}) = P(\text{White balls}) + P(\text{Red SuperBall}) = \frac{34}{55} + \frac{1}{8}\]

 

\[= \frac{272}{440} + \frac{55}{440}\]

 

\[= \frac{272 + 55}{440}\]

 

\[= \frac{327}{440}\]

 

So, the probability of winning a super prize in the SuperLottery is \( \frac{327}{440} \).

09.05.2024