Hi69420

1. The equation is simplified to to find the enclosed area.
 

2. The equation is further rewritten in standard form as (x - 8)^2 + (y + 11)^2 = 271, representing a circle centered at (8, -11) with a radius of the square root of 271 units.

 3. The area enclosed by the circle is calculated using the formula A = 271π square units.

4. Therefore, the area of the region enclosed by the given graph is 271π square units.

Here is my attempt 

1. The leading term of f(x) * g(x) is -x^5.

2. The leading coefficient of f(x) * g(x) is -1.

3. The degree of f(x) * g(x) is 5.

4. The constant term of f(x) * g(x) is 0.

5. The coefficient of x^2 in f(x) * g(x) is 0.

1. The row in Pascal's triangle has entries 1, n, ..., n, 1 with an average of 2.
 

2. The sum of entries in the row is calculated as 2n.

3. In Pascal's triangle, each entry is the sum of the two entries above it in the previous row.

4. The equation 2n = 2^(n-1) is solved to find the value of n.

5. The value of n in the row of Pascal's triangle is determined to be 2.

To calculate the probability of winning a super prize in the SuperLottery, we need to consider the different ways in which a ticket can win. Here are the steps to solve this problem:

1. Determine the total number of possible outcomes when drawing the balls.

2. Calculate the number of ways to win a super prize: a. Match at least two white balls: Choose 2 out of 3 winning white balls and any 1 of the remaining 9 white balls. This can be calculated as (3 2)x(9 1) . b. Match the SuperBall: Choose 1 out of 1 SuperBall from the red balls. c. Match both the SuperBall and at least two white balls: This can be calculated as the product of the number of ways to win in (a) and (b).

3. Calculate the total number of favorable outcomes for winning a super prize by summing up the results from (a), (b), and (c).

4. Determine the total number of possible outcomes by considering all ways the balls can be drawn.

5. Finally, calculate the probability of winning a super prize by dividing the total number of favorable outcomes by the total number of possible outcomes. By following these steps, you can determine the probability of winning a super prize in the SuperLottery.

Here is my attempt 

To find the probability that point P is closer to point O than to any of the side lengths in equilateral triangle XYZ, we need to understand the geometry involved. Here are the steps to solve this problem:

1. Draw equilateral triangle XYZ with center O.

2. The region closer to O than any of the side lengths forms an equilateral triangle inside triangle XYZ. Let's call this inner equilateral triangle LMN.

3. The area closer to the side lengths than to O consists of three smaller equilateral triangles that are outside the inner equilateral triangle.

4. Calculate the ratio of the area of the inner equilateral triangle to the total area of triangle XYZ. This ratio will give us the probability that point P is closer to O than to any of the side lengths.

5. The ratio of the areas of two similar figures is equal to the square of the ratio of their corresponding sides. Since the side length of the inner equilateral triangle is half that of triangle XYZ, the area ratio is (1/2)^2 = 1/4.

6. Therefore, the probability that a randomly chosen point P is closer to O than to any of the side lengths is 1/4 or 25%. This explanation should help you understand how to calculate the probability in this scenario.

Here is my attempt.
 

1. Given the equation "tan x + sec x = 0", we need to find the value of cos x.
 

2. Rewrite the expression as "(sin x + 1) / cos x = 0".
 

3. This implies that sin x = -1, indicating that x is in the fourth quadrant.

4. Using the Pythagorean identity, we find that cos x = 0 for an acute angle satisfying the equation.

Here is my attempt.
To simplify the expression \(\dfrac{1}{\sqrt2+\sqrt3}+\dfrac{1}{\sqrt2-\sqrt3}\), we need to rationalize the denominators of the fractions.
 

Let's start by rationalizing the first fraction: \(\dfrac{1}{\sqrt2+\sqrt3}\cdot \dfrac{\sqrt2-\sqrt3}{\sqrt2-\sqrt3}= \dfrac{\sqrt2-\sqrt3}{2-3}=-(\sqrt2-\sqrt3)=-\sqrt2+\sqrt3\)

Next, rationalize the second fraction: \(\dfrac{1}{\sqrt2-\sqrt3}\cdot \dfrac{\sqrt2+\sqrt3}{\sqrt2+\sqrt3}= \dfrac{\sqrt2+\sqrt3}{2-3}=-(\sqrt2+\sqrt3)=-\sqrt2-\sqrt3\)

Now, add the two rationalized fractions together:\(-\sqrt2+\sqrt3+(-\sqrt2-\sqrt3)=-\sqrt2+\sqrt3-\sqrt2-\sqrt3=-2\sqrt2\)

Therefore, the simplified expression is \(-2\sqrt2\).

To find the area of triangle ABC, we can first determine the height of the triangle from point A to side BC. Let's denote the height as h.

1. Let's use the formula for the area of a triangle: Area = 1/2 x base x height. 

2. Since we know AX = 10, BC = 10, and CX = 5, we can find the length of BX by subtracting CX from BC: BX=BC-CX = 10-5=5.

3. Now, let's consider triangle ABX and triangle ACX. Given that the circumcircles of these triangles have the same radius, we can deduce that triangles ABX and ACX are similar triangles. 4. By the property of similar triangles, the ratio of the sides of the two triangles is equal: 5. Since AX = 10, CX = 5, and BC = 10, we can find AB using the ratio: AX/AB = CX/AC = BX/BC Solving for AB gives: AB=2xAX=2x10=20

6. Now that we have AB = 20 and BC = 10, we can use these side lengths to find the height of triangle ABC.
 

7. To find the height h, we can use the Pythagorean theorem: Solving for h gives:\(h=\sqrt{400-25}=\sqrt{375}=5\sqrt3\)

8. Finally, we can calculate the area of triangle ABC using the formula we mentioned earlier: Therefore, the area of triangle ABC is Area= 1/2 x BC x h= 1/2x10x\(5\sqrt3=25\sqrt3\)

20.6, 79.7,79.7 degrees