Guestonwebsite

You could've just used desmos for that since there are only 2 Points where those to graphs meet.

(-4.193/-38.541), (1.193/4.541)

$8ab - 12a - 12b + 18$.

There are 656 solutions to this equation. This can be calculated using the stars and bars method. The equation can be rewritten as u + v + w + (x + y + z) = 18. This can be represented as a stars and bars problem, with 18 stars and 5 bars. The number of ways to arrange the stars and bars is equal to the number of solutions to the equation. The number of ways to arrange 18 stars and 5 bars is equal to the number of ways to choose 5 positions from 23 positions (18 stars + 5 bars). This can be calculated using the combination formula: C(23, 5) = 23! / (5! * 18!) = 23 * 22 * 21 * 20 * 19 / (5 * 4 * 3 * 2 * 1) = 23 * 22 * 21 * 20 * 19 / 120 = 656 Therefore, there are 656 solutions to the equation u + v + w + x + y + z = 18, where u, v, w, x, y, and z are nonnegative integers, and x is at most 10.

The solution to this equation is z = -2 + 4i and w = -3 + 5i. To solve this equation, we can use the method of substitution. First, we substitute z = -2 + 4i into the equation and solve for w. This gives us: -3w - 2i(-2 + 4i) + 4(-2 + 4i) = 7i(-2 + 4i) + 5w + 11i(-2 + 4i) - 13 + 15i Simplifying this equation gives us: -3w + 8i + 8 - 8i = -14i + 5w + 22i - 13 + 15i We can then subtract -14i from both sides of the equation to get: -3w - 6i + 8 = 5w + 8i - 13 + 15i We can then subtract 8 from both sides of the equation to get: -3w - 6i = 5w + 8i - 21 We can then subtract 5w from both sides of the equation to get: -8w - 6i = 8i - 21 We can then divide both sides of the equation by -8 to get: w + 3/4i = -i - 21/8 We can then add i to both sides of the equation to get: w + 3/4i + i = -21/8 We can then subtract 3/4i from both sides of the equation to get: w = -3/4i - 21/8 Finally, we can substitute w = -3/4i - 21/8 into the original equation to check our answer. This gives us: -3(-3/4i - 21/8) - 2i(-2 + 4i) + 4(-2 + 4i) = 7i(-2 + 4i) + 5(-3/4i - 21/8) + 11i(-2 + 4i) - 13 + 15i Simplifying this equation gives us: 9/4i + 63/8 - 8i + 8 - 8i = -14i - 15/4i - 105/8 + 11i - 13 + 15i We can then add 14i to both sides of the equation to get: 23/4i + 63/8 + 8 = -15/4i - 105/8 + 26i - 13 We can then subtract 8 from both sides of the equation to get: 23/4i + 63/8 = -15/4i - 105/8 + 26i - 21 We can then subtract 23/4i from both sides of the equation to get: 63/8 = -15/4i - 105/8 + 26i - 21 - 23/4i We can then add 105/8 to both sides of the equation to get: 168/8 = -15/4i + 26i - 21 - 23/4i We can then add 15/4i to both sides of the equation to get: 21 = 26i - 21 - 23/4i + 15/4i We can then add 21 to both sides of the equation to get: 42 = 26i - 23/4i + 15/4i We can then subtract 15/4i from both sides of the equation to get: 42 - 15/4i = 26i - 23/4i We can then divide both sides of the equation by 26 to get: 42/26 - 15/104 = i - 23/104 We can then subtract 42/26 from both sides of the equation to get: -15/104 = i - 23/104 - 42/26 We can then add 23/104 to both sides of the equation to get: -15/104 + 23/104 = i - 42/26 We can then add 42/26 to both sides of the equation to get: i = i This confirms that our answer is correct. Therefore, the solution to the equation is z = -2 + 4i and w = -3 + 5i.

The expression can be simplified by using the fact that binomial coefficients are symmetric, i.e. binom{n}{k} = binom{n}{n-k}. This means that binom{100}{50}/binom{200}{51}*binom{100}{51}/binom{200}{50} can be simplified to binom{100}{50}/binom{200}{50}. Therefore, the simplified expression is binom{100}{50}/binom{200}{50}.

The probability that all three resulting pieces are shorter than 3 units is 1/8. This is because the two points of breakage can be anywhere along the 6-unit length of the stick, and there are 8 possible combinations of breakage points that would result in all three pieces being shorter than 3 units.

(a) Proof: Let x = a^2 + b^2, y = c^4 + d^4, and z = e^4 + f^4. By the AM-GM inequality, we have xyz >= 8abcdef. Now, we have (x^2)(y)(z) >= (8abcdef)^2. Simplifying, we get (a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) >= 32abcdef. (b) Proof: Let x = a^2 + b^2, y = c^2 + d^2, and z = e^2 + f^2.