bader

We can solve the equation for u and v first:

Combine like terms: 7x^2 + 2x - 21 = 0

Factor the equation: (x - 3)(7x + 7) = 0

Therefore, the solutions are x = 3 and x = -1. These correspond to u = 3 and v = -1.

Now, we need to find 1/u^2 + 1/v^2:

Substitute u = 3 and v = -1: 1/3^2 + 1/(-1)^2

Simplify: (1/9) + (1) = 1/9 + 9/9

Combine terms: 10/9

Therefore, 1/u^2 + 1/v^2 = 10/9.

There are two cases to consider for placing the two indistinguishable pieces on the chessboard:

Same Row: The pieces can be placed on any of the 8 rows. Once the row is chosen, either piece can be placed on any of the 8 squares in that row.

So, for the same row scenario, there are 8 ways to pick a row and 8 ways to arrange the pieces within that row, resulting in 8 * 8 = 64 arrangements.

Same Column (excluding diagonal): The pieces can be placed on any of the 8 columns (excluding the possibility of them being on the same diagonal, which we'll address later).

Similar to the same row case, once a column is chosen, either piece can be placed on any of the 8 squares in that column.

However, this counts the diagonal placement twice (once for each way of choosing the row or column first).

So, for the same column scenario (excluding diagonal), there are 7 ways to pick a column and 7 ways to arrange the pieces within that column, resulting in 7 * 7 = 49 arrangements.

Correcting for Diagonal Placement:

We've double-counted the arrangements where the pieces are on the same diagonal.

How many are there? Imagine choosing the diagonal first. There are only 4 choices (one for each corner).

Then, you can place the pieces on two squares out of the 8 on that diagonal, giving 8 choices.

However, since we've already counted this scenario twice (once for each way of choosing the row or column first), we need to subtract 1.

Therefore, the number of diagonal placements counted twice is 4 (diagonals) * (8C2) (combinations to choose 2 squares out of 8) - 1 (double counting) = 14.

Total Arrangements:

Adding the arrangements for same row, same column (excluding diagonal), and subtracting the double-counted diagonals, we get the total number of possible placements:

Total = Same Row + Same Column (excluding diagonal) - Double Counted Diagonals Total = 64 + 49 - 14 Total = 99

There are 99 ways to place two indistinguishable pieces on an 8x8 chessboard if they must be in the same row or the same column.

To compute \( \log_{8^a} 4^b \) in terms of \( a \) and \( b \), we'll utilize the properties of logarithms.

First, let's express \( 8^a \) and \( 4^b \) in terms of a common base. Since both 8 and 4 can be represented as powers of 2, we rewrite them as follows:

\[ 8^a = (2^3)^a = 2^{3a} \]

\[ 4^b = (2^2)^b = 2^{2b} \]

Now, our expression becomes:

\[ \log_{8^a} 4^b = \log_{2^{3a}} 2^{2b} \]

Using the property of logarithms that \( \log_a b^n = n \cdot \log_a b \), we rewrite this as:

\[ = 2b \cdot \log_{2} 2^{3a} \]

Since \( \log_{2} 2^{3a} = 3a \) (as \( \log_{a} a = 1 \)), our expression simplifies to:

\[ = 2b \cdot 3a \]

Thus, \( \log_{8^a} 4^b \) in terms of \( a \) and \( b \) is \( 6ab \).

Let O be the center of the inscribed circle. Let E be the midpoint of AB, and let F be the midpoint of CD. Draw segments AO, BO, CO, and DO. Since ABCD is an isosceles trapezoid, ∠ABE=∠CBE and ∠DCE=∠EDF, so triangles ABE, CBE, DCE, and EDF are all isosceles.

Since AE=BE, CE=DE, and AO=BO, triangles AEO, BEO, CEO, and DEO are all congruent. Therefore, OA=OB=OC=OD=r, where r is the radius of the inscribed circle.

Let E′ be the foot of the altitude from E to AB, and let F′ be the foot of the altitude from F to CD. Since triangles AEE′ and BEE′ are both right isosceles triangles, AE′=BE′=r, and similarly, CF′=DF′=r. Therefore, the sum of the lengths of the segments from E to D is: [ EE' + ED + DF' = 2r + x + 2r = x + 4r. ]

Similarly, the sum of the lengths of the segments from F to B is y+4r. Since ABCD is an isosceles trapezoid, the sum of the lengths of the segments from E to D is equal to the sum of the lengths of the segments from F to B. Therefore, x+4r=y+4r, so x=y. Hence, the radius of the inscribed circle is sqrt(xy)​​.

The solid formed by rotating triangle ABC around side BC is a cone. Here's how to find its volume:

Identify the cone's dimensions:

Radius (r): During the rotation, side AC traces out the base of the cone

 Since AC = 5, the base radius (r) of the cone is 5/2 (half the length of AC becomes the radius).

Height (h): As the triangle rotates, side AB acts as the height of the cone. Therefore, the height (h) of the cone is 5 (AB length).

Formula for Cone Volume: The volume of a cone is calculated using the formula: Volume = (1/3) * π * r^2 * h

Substitute Values and Solve: Volume = (1/3) * π * [(5/2)^2] * 5 Volume = (1/3) * π * (25/4) * 5 Volume = (125/12) * π

Answer: The volume of the solid is (125/12)π cubic units.

To evaluate the expression f(1) + f(2) + f(3) + ... + f(999) + f(1000), we need to understand the behavior of the function f(x) first.

The function f(x) is defined as the floor of (2 - 3x)/(x + 3). The floor function rounds down to the nearest integer. We can observe the behavior of this function for different values of x to understand how to proceed.

Let's analyze the function for a few values of x:

f(1) = floor((-1)/4) = -1

f(2) = floor((-4)/5) = -1

f(3) = floor((-7)/6) = -2

f(4) = floor((-10)/7) = -2

f(5) = floor((-13)/8) = -2

It seems that f(x) remains constant within intervals of x before decreasing by 1 and remaining constant again.

Now let's find the length of each interval:

For x = 1 to x = 4, f(x) = -1.

For x = 5 to x = 8, f(x) = -2.

This pattern continues. So, for each interval of length 4, the value of f(x) remains constant, then decreases by 1.

The sum of all f(x) from x = 1 to x = 1000 can be calculated by dividing 1000 by 4 to find out how many complete cycles occur, and then multiplying by the sum of each cycle plus the remaining values.

Number of complete cycles: 1000/4 = 250.

In each cycle:

For x = 1 to x = 4, the sum is -1 -1 -1 -1 = -4.

For x = 5 to x = 8, the sum is -2 -2 -2 -2 = -8.

So, the sum of each complete cycle is -4 -8 = -12.

The remaining values are f(997), f(998), f(999), f(1000). These are -3, -3, -3, -3 respectively.

So, the total sum is 250 * (-12) - 12 = -3000 - 12 = -3012.

When I graphed it, I got f(-2) = 1/2.  Hope this helps!

There are 2 ways to start with the letter A: either from the first A in the array, or from the second A.

To reach the second A, we must move down once. Once we have reached the first or second A, we can move in any direction to reach the letter R.

There are 3 ways to do this: we can move down, right, or down and then right.

Once we have reached the letter R, we can again move in any direction to reach the letter C.

There are 3 ways to do this: we can move down, left, or left and then down.

Once we have reached the letter C, the last step is to move up to reach the letter H, and there is only 1 way to do this.

Therefore there are 2×3×3×1 = 18​ ways to spell the word "ARCH" in the array provided.

First, we can rewrite the numerator:

lim theta -> 0 (1 - cos theta)/theta = lim theta -> 0 2 sin(theta/2)^2/theta

We can then simplify and apply l'Hopital's Rule:

lim theta -> 0 2 sin(theta/2)^2/theta = lim theta ->0  4 sin (theta/2) cos(theta/2)/1 = 4.

Therefore, the limit is 4.