nathanl6656

Let's visualize the problem:

Imagine a right-angled corner of a room. Place a circular table in this corner so that one edge of the table touches both walls.

Point P: A point on the edge of the table that is 8 units away from one wall and 9 units away from the other.

Radius: The distance from the center of the table to any point on its edge (like point P).

Solution:

Form a right triangle:

The distance from point P to the first wall (8 units) is one leg of a right triangle.

The distance from point P to the second wall (9 units) is the other leg of the right triangle.

The hypotenuse of this right triangle is the diameter of the table, which is twice the radius.

Use the Pythagorean theorem:

a² + b² = c²

8² + 9² = c²

64 + 81 = c²

145 = c²

c = √145

Find the radius:

Since the diameter is twice the radius, the radius is half of √145.

Radius = (√145) / 2

Therefore, the radius of the table is (√145) / 2 units.

The answer is 4n + 3.

Finding the Area of Triangle XYZ

The area of a right triangle is (base * height)/2.

In this case, base = XY = 12 and height = YZ = 10.

So, area of triangle XYZ = (12 * 10)/2 = 60.

Relating the Area of XYD to the Height from D to XY

Let h be the height of triangle XYD from point D to side XY.

The area of triangle XYD = (base * height)/2 = (12 * h)/2 = 6h.

Finding the Maximum Height for an Area of 24

We want the area of XYD to be at most 24.

So, 6h <= 24.

Solving for h, we get h <= 4.

Calculating the Probability

The probability of choosing a point D with a height less than or equal to 4 is equal to the ratio of the area where the height is less than or equal to 4 to the total area of triangle XYZ.

If we draw a line parallel to XY at a distance of 4 units from XY, it will divide the triangle into two similar triangles. The smaller triangle will have a height of 4 and its area will be (1/4) * (area of triangle XYZ) = (1/4) * 60 = 15.

So, the probability of choosing a point D with a height less than or equal to 4 (and thus an area of XYD at most 24) is 15/60 = 1/4.

Therefore, the probability that the area of triangle XYD is at most 24 is 1/4.

The area of OPQ is 16/3.

1. The largest possible value of |a_1 - a_2| is 20.

2. The largest possible value of |a_1 - a_2| is 10.

(a) Let's first find out how much magical syrup is in 300 mL of blue potion. Since the blue potion is 15% magical syrup by volume, the amount of magical syrup in 300 mL of blue potion is 0.15 x 300 = 45 mL.

Now, let's assume that we add x mL of red potion to this mixture. The resulting potion will have a total volume of 300 + x mL, and its magical syrup content will be (45 mL + 0.5x mL) / (300 mL + x mL). We want this to be equal to 20% (or 0.2), so we can set up the following equation:

(45 mL + 0.5x mL) / (300 mL + x mL) = 0.2

Solving for x, we get:

x = 150 mL

Therefore, 150 mL of red potion must be added to 300 mL of blue potion to produce a potion that is 20% magical syrup by volume.

(b) Let's assume that we mix x mL of red potion with y mL of blue potion to produce 180 mL of a potion that is 40% magical syrup by volume. We can set up the following two equations based on the amount of magical syrup and the total volume:

Amount of magical syrup: 0.5x mL + 0.15y mL = 0.4(180 mL)

Total volume: x mL + y mL = 180 mL

Simplifying the first equation, we get:

0.5x + 0.15y = 72

We can now use substitution to solve for x and y. Solving the second equation for y, we get:

y = 180 - x

Substituting this into the first equation, we get:

0.5x + 0.15(180 - x) = 72

Solving for x, we get:

x = 120 mL

Substituting this into the equation y = 180 - x, we get:

y = 60 mL

Therefore, 120 mL of red potion and 60 mL of blue potion can be combined to produce 180 mL of a potion that is 40% magical syrup by volume.

(c) Let's assume that we mix x mL of red potion with y mL of blue potion to produce a potion that is 25% magical syrup by volume. We can set up the following two equations based on the amount of magical syrup and the total volume:

Amount of magical syrup: 0.5x mL + 0.15y mL = 0.25(x+y) mL

Total volume: x mL + y mL = some value

We have two equations and two unknowns, so we can solve for x and y. Simplifying the first equation, we get:

0.25x - 0.1y = 0

Substituting the second equation into this equation, we get:

0.25x - 0.1(x + y) = 0

Simplifying, we get:

0.15x - 0.1y = 0

We can now use substitution to solve for x and y. Solving the second equation for y, we get:

y = some value - x

Substituting this into the first equation, we get:

0.15x - 0.1(some value - x) = 0

Simplifying, we get:

0.25x - 0.1(some value) = 0

Therefore, we can see that there is no solution that satisfies these equations for any positive values of x and y. So there is no combination of red

The value of b is 3*2^(3/4).

The possible values of a are -35, -12, -10, 10, 12, and 35.

Here are the answers:
\[\mathbf{A}^{-1} \mathbf{w} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}\]

\[\mathbf{A}^{-1} \mathbf{x} = \begin{pmatrix} 4 \\ 0 \end{pmatrix}\]

\[\mathbf{A}^{-1} \mathbf{y} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\]

\[\mathbf{A}^{-1} \mathbf{z} = \begin{pmatrix} 5 \\ 1 \end{pmatrix}\]

AB + AC + BC = 28

To solve this problem, let's first consider the structure of the cube.

When the large cube is cut into \(6^3\) smaller cubes, each small cube will have one face painted black and the remaining faces unpainted (since each face of the large cube was painted black initially).

Now, if we roll one of these small cubes, regardless of which face is on top initially, it has a \(1/6\) chance of landing with each face facing upward, since all faces are identical.

However, the probability that the face on top after rolling is black depends on whether the cube was rolled along an axis perpendicular to a black face or not. If it was, the black face will still be on top; if not, it won't be.

Considering that the black faces are on the outside of the original cube, any roll along the x, y, or z-axis will result in the black face being on top. There are 3 such axes.

So, the probability of the black face being on top after rolling is:

\[P(\text{Black face on top}) = \frac{3}{6} = \frac{1}{2}\]

Hence, the probability that when the small cube is rolled, the face on the top is black is \( \frac{1}{2} \).