All Answers 355864 Answers

             A    E

B                 D                  C  15

      10                  10   

EC / sin 90 = DC / sin75

EC = 10/sin75  

BD = DC

ED = ED

So....by LL,  right triangle BED  congruent to right triangle CED

So BE  = 10/sin75 ≈ 10.35

nvm I got the answer, 

Since \(BQ = \frac{1}{3} AB\), \(BR = \frac{1}{3} BC\), and \(\angle QBR = \angle ABC\), triangles QBR and ABC are SAS-similar. Furthermore, since Q and R are trisection points, the side lengths are in a \(1:3\) ratio, so the areas are in a \(1:9\) ratio. This gives \([QBR] = \frac{1}{9} [ABC].\) 

Likewise, triangles UDV and CDA are similar, and \([UDV] = \frac{1}{9} [CDA].\)

Therefore, \([QBR] + [UDV] = \frac{1}{9} ([ABC] + [CDA]) = \frac{1}{9} [ABCD] = 20.\)

The remainder of quadrilateral ABCD is hexagon \(AQRCUV\), so it has area \(180 - 20 = \boxed{160}.\)

sqrt (3) / 3 = (1/2) s^2 * sqrt (3) / 2       where s is the side of the triangle

1/3 = (1/4)s^2

4/3  = s^2

We have triangle APB

angle PAB = angle PBA = 30

angle APB = 120

Law of Cosines

s^2 = 2x^2 - 2x^2 * cos (120)

s^2 = 2x^2 - 2x^2 *(-1/2)

s^2 = 3x^2

4/3 = 3x^2

4/9 = x^2

2/3 = x

Gang they asked for hints XD

a) 18^2 * (-4)^2 and (49 + 121)(49 - 121) => 5184 and -12240

b) (x+y)(x-y)(x+y)(x-y) = (x^2 - y^2)^2.

When does (x^2 - y^2)^2 = (x^2 + y^2)(x^2 - y^2)? Divide both sides by (x^2 - y^2) => x^2 - y^2 = x^2 + y^2

Subtract x^2 from both sides, you get -y^2 = y^2, add y^2 to both sides, 2y^2 = 0, y^2 = 0, y = 0. Thus the two expressions are equal when y = 0, and x = whatever. 

For which values of x and y does (x+y)62(x-y)^2 not equal (x^2 + y^2)(x^2 - y^2)? It happens when y is not 0, and x is whatever.

a) 18^2 * (-4)^2 and (49 + 121)(49 - 121) => 5184 and -12240

b) (x+y)(x-y)(x+y)(x-y) = (x^2 - y^2)^2.

When does (x^2 - y^2)^2 = (x^2 + y^2)(x^2 - y^2)? Divide both sides by (x^2 - y^2) => x^2 - y^2 = x^2 + y^2

Subtract x^2 from both sides, you get -y^2 = y^2, add y^2 to both sides, 2y^2 = 0, y^2 = 0, y = 0. Thus the two expressions are equal when y = 0, and x = whatever. 

For which values of x and y does (x+y)62(x-y)^2 not equal (x^2 + y^2)(x^2 - y^2)? It happens when y is not 0, and x is whatever.

Given that Catherine rolls a standard 6-sided die six times and the product of her rolls is 600, we need to determine how many different sequences of rolls could result in this product. 

Firstly, we factorize 600 into its prime factors:
\[ 600 = 2^3 \times 3 \times 5^2 \]

The numbers on a 6-sided die are 1, 2, 3, 4, 5, and 6. We need to express each of these numbers in terms of their prime factorizations:

- \(1 = 1\)
- \(2 = 2\)
- \(3 = 3\)
- \(4 = 2^2\)
- \(5 = 5\)
- \(6 = 2 \times 3\)

The product of these rolls must equate to \(2^3 \times 3 \times 5^2\). Since each roll can only be 1, 2, 3, 4, 5, or 6, we must distribute the prime factors accordingly among the six rolls.

### Step 1: Distribute the Prime Factors

Each roll will contribute to the overall prime factorization of 600. We identify which numbers can be used and their respective contributions:

- \(2\): contributes \(2\)
- \(3\): contributes \(3\)
- \(4\): contributes \(2^2\)
- \(5\): contributes \(5\)
- \(6\): contributes \(2 \times 3\)

### Step 2: Determine Possible Rolls

To satisfy the factorization \(2^3 \times 3 \times 5^2\) with six rolls:

- We need three 2's.
- We need one 3.
- We need two 5's.

Possible combinations of rolls can be:

- 2 can be contributed by 2, 4 (as \(2^2\)), or 6 (as \(2\)).
- 3 can be contributed by 3 or 6 (as \(3\)).
- 5 can be contributed by 5.

### Step 3: Identify Suitable Rolls

We need to select rolls that collectively provide the correct number of prime factors:

- \(2\) appears in rolls: \(2, 4, 6\).
- \(3\) appears in rolls: \(3, 6\).
- \(5\) appears in rolls: \(5\).

### Step 4: Determine Combinations

Let's break down how we can achieve the required distribution:

- \(2^3\) can be achieved by: 
  - (2, 2, 2)
  - (4, 2)
  - (4, 4)
  - (2, 6)
  - (6, 6)
  
- \(3\) must be from: 
  - (3)
  - (6)
  
- \(5^2\) must be from: 
  - (5, 5)

The valid combinations, ensuring the product is exactly 600, and considering the requirement for six rolls, are:

\[ (2, 2, 2, 3, 5, 5) \]

### Step 5: Counting Permutations

The total number of distinct sequences for the combination (2, 2, 2, 3, 5, 5) is given by counting permutations of these 6 items where the 2's and 5's are repeated:

\[ \frac{6!}{3! \cdot 2!} = \frac{720}{6 \cdot 2} = 60 \]

Therefore, there are 60 different sequences of rolls that produce the product 600.

Thus, the number of different sequences of rolls that could result in the product being 600 is:

\[ \boxed{60} \]

To solve the problem of finding the number of ways the numbers 1 through 5 can be arranged into the five boxes so that the inequalities \(\_ < \_ > \_ < \_ > \_\) hold, we need to carefully consider the pattern and constraints.

Let's break it down step by step:

1. **Pattern Analysis:**

   - The sequence alternates between increasing and decreasing.

   - If we let the positions be labeled as \(a < b > c < d > e\), we need to ensure that each set of inequalities is satisfied.

2. **Key Points:**

   - The first number must be less than the second number.

   - The second number must be greater than the third number.

   - The third number must be less than the fourth number.

   - The fourth number must be greater than the fifth number.

3. **Counting Valid Sequences:**

   - There are a total of 5! (120) possible permutations of the numbers 1 through 5.

   - We need to determine how many of these permutations satisfy the given inequalities.

One way to approach this problem is to use a systematic method to generate valid sequences, but a more efficient way is to realize that this problem is a known combinatorial problem.

### Generating Function Approach (Optional):

This problem can also be solved using generating functions and other combinatorial techniques, but for simplicity, let's use a direct combinatorial argument.

### Direct Counting:

Instead of listing all permutations, we can use a combinatorial argument based on symmetry and known results in permutations and inequalities.

For a permutation of \(n\) elements satisfying a specific pattern of inequalities like this, it can be shown that the number of valid permutations is given by:

\[ \frac{(n-1)!}{(k-1)! \cdot (n-k-1)!} \]

where \(k\) is the number of ascents (increasing steps) in the permutation.

For the pattern \(_ < _ > _ < _ > _\), there are 2 ascents and 2 descents in the sequence.

Therefore, the number of valid permutations is:

\[ \frac{4!}{1! \cdot 1! \cdot 2!} = \frac{24}{2} = 12 \]

Thus, there are 12 valid ways to arrange the numbers 1 through 5 into the five boxes to satisfy the given inequalities.

The answer is \(48\)

The factors of 48 are \(1, 2, 3, 4, 6, 8, 12, 16, 24, 48\)

There are exactly 10. 

Thanks! :)

Let's look at the number 512. 

We know that \(512=2^9\) , which means it has 10 divisors. 
Since it's divisble by 2 and two is the smallest prime, 

our final answer is 2. 

Thanks! :)

HOW IS THIS POSSIBLE? 

Look at the image. How can every scientist sit next to a mathematician?

Let's calculate the number of ways we can seat the children and adults seperately. 

First, there are \(4! = 24\)  ways to seat the children and  \(4! = 24\) ways to seat the adults. 

We now multiply these values together to get

\(24*24=576\)

So 576 is our answer. 

Thanks! :)

Multiply both sides by 3

3xy + 6x  = x - 2y + z

3xy + 5x = z - 2y

x ( 3y + 5)  = z - 2y

x = [ z -2y ] / [ 3y + 5]

Here's my attempt

The total number of ways of choosing  5 cards from  28 = C(28,5)

Out of  any of the 7 numbers choose any  1   = C(7,1)

We have 4 choices of the same  number, choose any 2 = C(4,2)

Out of the other 6 numbers, choose any 3 = C(6,3)

In  each of these choices, we can  choose any 1 of the 4 numbers = [ C(4,1)]^3 = 4^3

The probability is

[ C (7,1) * C(4,2) * C (6,3) * 4^3 ]  / C ( 28,5)  = 64 /117  ≈  54.7 %  { as Ishiaki found !!! }

: | that was the answer but ok?

Let's solve for x in terms of y and z:

1. Combine like terms with x:

xy + 2x - x = -2y + z/3

2. Simplify the equation:

(x + xy) - x = -2y + z/3

xy = -2y + z/3

3. Isolate x:

Divide both sides by y (assuming y ≠ 0). Note that if y = 0, there would be no solution for x, as the denominator would be zero.

Therefore, x = (-2y + z/3) / y

Simplified form: We can further simplify the expression by combining terms in the numerator:

x = (-6 + z) / (3y)

This form expresses x in terms of y and z.

I solved it!

 

Given:

α lies in quadrant II, and tan(α) = -12/5.

β lies in quadrant IV, and cos(β) = 3/5.

We need to find sin(α + β).

Step 1: Determine sin(α) and cos(α)

Use the given value of tan(α): tan(α) = -12/5

Use the identity tan(α) = sin(α)/cos(α): tan(α) = sin(α)/cos(α)

Assume sin(α) = 12k and cos(α) = 5k, where k is a constant.

Apply the Pythagorean identity sin²(α) + cos²(α) = 1: (12k)² + (5k)² = 1 → 144k² + 25k² = 1 → 169k² = 1 → k² = 1/169 → k = 1/13

Substitute k back to find sin(α) and cos(α): sin(α) = 12k = 12/13, cos(α) = 5k = 5/13

Since α is in quadrant II, cos(α) should be negative: cos(α) = -5/13

Step 2: Determine sin(β) from cos(β)

Use the given value of cos(β): cos(β) = 3/5

Apply the Pythagorean identity sin²(β) + cos²(β) = 1: sin²(β) + (3/5)² = 1 → sin²(β) + 9/25 = 1 → sin²(β) = 1 - 9/25 = 25/25 - 9/25 = 16/25

Take the square root to find sin(β): sin(β) = ±4/5

Since β is in quadrant IV, sin(β) should be negative: sin(β) = -4/5

Step 3: Use the angle sum formula for sine

Use the angle sum formula: sin(α + β) = sin(α) cos(β) + cos(α) sin(β)

Substitute the known values of sin(α), cos(α), sin(β), and cos(β): sin(α + β) = (12/13)(3/5) + (-5/13)(-4/5)

Simplify each term: sin(α + β) = (12 * 3) / (13 * 5) + (5 * 4) / (13 * 5) = 36/65 + 20/65

Add the fractions: sin(α + β) = (36 + 20) / 65 = 56/65

Therefore, the exact value of sin(α + β) is: 56/65

1. Total number of ways to choose 5 cards from 28:

   C(28, 5) = 28! / (5! * 23!) = 98280
 

2. Number of ways to draw 5 cards such that exactly two have the same number:
   - Choose 1 number to appear twice: 

   C(7, 1) = 7

   - Choose 2 out of the 4 available cards for this number: 

   C(4, 2) = 6

   - Choose 3 different numbers from the remaining 6: 

   C(6, 3) = 20

   - Choose 1 card for each of these 3 numbers from the 4 available: 

   4^3 = 64
 

3. Calculate the total number of favorable outcomes:

   7 * 6 * 20 * 64 = 53760
 

4. Calculate the probability:

   Probability = 53760 / 98280 approximately 0.547

Final Answer

The probability that exactly two of the five cards Yunseol draws have the same number is approximately 0.547 or 54.7%.

(That should be the correct answer)

Greetings Answerscorrectly, 

I assume you have at least been learning something from your course/Alcumus, so I'll keep the hints short.

To start, we want to create a fraction something like this:
 

\(\frac{\textbf{Number of Targeted Results}}{\textbf{Number of All Results}}\)

To find the numerator, we want to find the probability for getting the desired card for each of the 5 moves.

If you don't know how to do that, here are instructions for each card draft.

1. Pick a random card regardless of colour and number

2. Pick a randomly coloured card that has the same number as 1.

3. Pick a randomly coloured card that is not the same number as 1. or 2.

4. Pick a randomly coloured card that is not the same number as 1., 2., or 3.

5. Pick a randomly coloured card that is not the same number as 1., 2., 3., or 4.

As for the denominator, I think you can figure that out yourself. 

Don't forget that cards are NOT put back into the pool after taken out!

If you need any more help or have found something wrong with my answer (hint?), say the word.

I hope this helps you answer correctly, Answerscorrectly! (heh, I've been waiting to say that for a while..)

Also just a disclaimer. although Idk how to delete answers, If I get the answer to this question I'm going to find a way on how to remove that cause oh man, I've seen soooo many ppl in my class just come here and read off the answers. so like just don't.