RedDragonl

To solve the given system of equations

\[
\frac{xy}{x + y} = 1, \quad \frac{xz}{x + z} = 2, \quad \frac{yz}{y + z} = 4,
\]

we first rewrite each equation in terms of its cross-multiplication:

1. \( \frac{xy}{x + y} = 1 \)

\[
xy = x + y
\]

\[
xy - x - y = 0 \quad \Rightarrow \quad xy - x - y + 1 = 1
\]

\[
(x-1)(y-1) = 2
\]

2. \( \frac{xz}{x + z} = 2 \)

\[
xz = 2(x + z)
\]

\[
xz - 2x - 2z = 0 \quad \Rightarrow \quad xz - 2x - 2z + 4 = 4
\]

\[
(x-2)(z-2) = 8
\]

3. \( \frac{yz}{y + z} = 4 \)

\[
yz = 4(y + z)
\]

\[
yz - 4y - 4z = 0 \quad \Rightarrow \quad yz - 4y - 4z + 16 = 16
\]

\[
(y-4)(z-4) = 20
\]

We now have three transformed equations:

\[
(x-1)(y-1) = 2
\]

\[
(x-2)(z-2) = 8
\]

\[
(y-4)(z-4) = 20
\]

We solve these equations step by step. Start with the first equation:

\[
(x-1)(y-1) = 2
\]

Express \( y \) in terms of \( x \):

\[
y = \frac{2}{x-1} + 1
\]

Next, substitute \( y \) into the second equation:

\[
(x-2)(z-2) = 8
\]

Express \( z \) in terms of \( x \):

\[
z = \frac{8}{x-2} + 2
\]

Substitute both \( y \) and \( z \) into the third equation:

\[
(y-4)(z-4) = 20
\]

Substituting \( y = \frac{2}{x-1} + 1 \) and \( z = \frac{8}{x-2} + 2 \):

\[
\left( \frac{2}{x-1} + 1 - 4 \right) \left( \frac{8}{x-2} + 2 - 4 \right) = 20
\]

Simplify \( y - 4 \) and \( z - 4 \):

\[
y - 4 = \frac{2}{x-1} - 3
\]

\[
z - 4 = \frac{8}{x-2} - 2
\]

So, the equation becomes:

\[
\left( \frac{2 - 3(x-1)}{x-1} \right) \left( \frac{8 - 2(x-2)}{x-2} \right) = 20
\]

Simplify the terms inside the parentheses:

\[
\left( \frac{2 - 3x + 3}{x-1} \right) \left( \frac{8 - 2x + 4}{x-2} \right) = 20
\]

\[
\left( \frac{5 - 3x}{x-1} \right) \left( \frac{12 - 2x}{x-2} \right) = 20
\]

Solve for \( x \). By testing rational values, we find:

Let \( x = 3 \):

\[
y = \frac{2}{3-1} + 1 = 2
\]

\[
z = \frac{8}{3-2} + 2 = 10
\]

Therefore, \(z = 10\).

The number of solutions is 3218.

The answer is 22/9.

The answer is 65/72.

1. The answer is 24 degrees.

2. The number of possible locations is 4.

3. x = 56 degrees.

The value of x is 3*sqrt(2).

The rope is 24 m long.

The answer is 32.

Problem 3:

To find the integers n that satisfy 64≤n2≤100, we can take the square root of both sides of the inequality. However, since the square root function is not one-to-one (it has two outputs for each positive input), we need to be careful.

Taking the square root of both sides gives:

8≤∣n∣≤10

Since we only care about positive integer solutions for n, we can simply remove the absolute value signs and consider 8≤n≤10.

Now, we can count the number of integers in this range, inclusive. There are 10−8+1=3​ integers that satisfy the inequality: 8, 9, and 10.

Problem 2:

To find the largest possible value of a + 2b + 3c, we want to maximize the values of a, b, and c.

Since a, b, and c can be the numbers 1, 2, 3 in some order, let's consider the case where a = 3, b = 2, and c = 1:

a + 2b + 3c = 3 + 2(2) + 3(1) = 3 + 4 + 3 = 10

Therefore, the largest possible value of a + 2b + 3c is 10.

Problem 1:

We can complete the square to minimize the given expression

Steps to solve:1. Complete the square

First, we recognize that the expression is already in the form of a perfect square trinomial up to a constant factor. This means we can complete the square by taking half of the coefficient of our x2 term, squaring it, and adding it to both sides of the equation.

In this case, the coefficient of our x2 term is −2, so half of it would be −1, and squaring it gives us 1. Because we're adding the constant term inside the parentheses on the right where it's being multiplied by 1 (the coefficient of x2 ), we need to add −1 (the opposite of half the coefficient of our x2 term squared) outside the parentheses on the left to make sure we're adding the same thing to both sides.

x4−2x2−1=x4−2x2−1

(x2)2−2(x2)+1=(x2−1)2

2. Minimum value:

Now that the expression is a squared term, we know its minimum value is 0, which occurs when x2=1 or x=±1.

Answer: The minimum value of the expression is x2(x2−2) is 0,​ which occurs when x=±1.

To find the distance between two opposite corners of the rectangular prism, we can visualize a diagonal line going through the entire prism. This diagonal will form a right triangle with sides that correspond to the lengths, width, and height of the prism.

We can use the Pythagorean Theorem to find the length of this diagonal.

Pythagorean Theorem:

a^2 + b^2 = c^2

where:

a and b are the lengths of the two legs of a right triangle.

c is the length of the hypotenuse (the diagonal in this case).

Applying the theorem:

In our case:

a = length of the prism = 3 units

b = width of the prism = 5 units

c = diagonal length (what we want to find)

Solve for c (diagonal length):

c^2 = a^2 + b^2

c^2 = 3^2 + 5^2

c^2 = 9 + 25

c^2 = 34

Taking the square root of both sides (remembering there might be positive and negative square roots), we get:

c = ±√34

Since c represents a distance (and cannot be negative), we take the positive square root:

c = √34 units

Therefore, the distance between two opposite corners of the rectangular prism is √34 units. (This is the exact answer, but you can approximate it to a decimal value if needed.)

To find the probability that intervals \(I\) and \(J\) overlap, we need to consider the possible positions of the four points \(x_1, x_2, x_3,\) and \(x_4\) within the interval \([0, 1]\).

Without loss of generality, let's assume that \(x_1 < x_2\) and \(x_3 < x_4\).

For \(I\) and \(J\) to overlap, one of the following conditions must be true:

1.  \(x_1 < x_3 < x_2 < x_4\)

2.  \(x_3 < x_1 < x_4 < x_2\)

Let's calculate the probability for each condition:

1.  Probability of Condition 1:

   - The probability that \(x_1\) falls in the interval \([0, 1]\) is \(1\).

   - The probability that \(x_3\) falls in the interval \([x_1, 1]\) is \(1 - x_1\).

   - Given \(x_1\) and \(x_3\), the probability that \(x_2\) falls in the interval \((x_1, 1]\) is \(1 - x_1\).

   - Given \(x_1\), \(x_3\), and \(x_2\), the probability that \(x_4\) falls in the interval \((x_3, 1]\) is \(1 - x_3\).

   - So, the probability of Condition 1 is \(1 \times (1 - x_1) \times (1 - x_1) \times (1 - x_3) = (1 - x_1)^2 (1 - x_3)\).

2.  Probability of Condition 2:

   - The probability that \(x_3\) falls in the interval \([0, 1]\) is \(1\).

   - The probability that \(x_1\) falls in the interval \([0, x_3]\) is \(x_3\).

   - Given \(x_3\) and \(x_1\), the probability that \(x_4\) falls in the interval \((x_3, 1]\) is \(1 - x_3\).

   - Given \(x_3\), \(x_1\), and \(x_4\), the probability that \(x_2\) falls in the interval \((x_1, 1]\) is \(1 - x_1\).

   - So, the probability of Condition 2 is \(1 \times x_3 \times (1 - x_1) \times (1 - x_3) = x_3 (1 - x_1) (1 - x_3)\).

Now, since \(x_1, x_2, x_3,\) and \(x_4\) are chosen independently and uniformly at random in the interval \([0, 1]\), we can find the probability that intervals \(I\) and \(J\) overlap by summing the probabilities of Condition 1 and Condition 2:

\[ \text{Total probability} = (1 - x_1)^2 (1 - x_3) + x_3 (1 - x_1) (1 - x_3) \]

\[ = (1 - x_1) (1 - x_3) (1 - x_1 + x_3) \]

\[ = (1 - x_1) (1 - x_3) \]

Now, since \(x_1\) and \(x_3\) are chosen uniformly at random in the interval \([0, 1]\), we can find the expected value of the probability by integrating over the joint distribution of \(x_1\) and \(x_3\):

\[ \text{Expected probability} = \int_{0}^{1} \int_{0}^{1} (1 - x_1) (1 - x_3) \, dx_1 \, dx_3 \]

\[ = \int_{0}^{1} \left( \int_{0}^{1} (1 - x_1) (1 - x_3) \, dx_3 \right) dx_1 \]

\[ = \int_{0}^{1} \left( 1 - x_1 - \frac{1}{2} + \frac{x_1^2}{2} \right) dx_1 \]

\[ = \left[ x_1 - \frac{x_1^2}{2} - x_1 + \frac{x_1^3}{6} \right]_{0}^{1} \]

\[ = \left( 1 - \frac{1}{2} - 1 + \frac{1}{6} \right) - \left( 0 - 0 - 0 + 0 \right) \]

\[ = \frac{1}{3} - \frac{1}{2} = \frac{1}{6} \]

Therefore, the expected probability that intervals \(I\) and \(J\) overlap is \( \frac{1}{6} \).