BuilderBoi

Case 1: Weeks 91 - 104

There are \(104 - 91 + 1 = 14\) weeks, so they produce a total of \(14 \times 1000 = 14000 \) toys. 

Case 2: Weeks 1 - 91

We have an arithmetic sequence with a starting term of 280, a final term of 100, and 91 terms.

This means that the sum is\((1000 + 280) \div 2 \times 91 = 58240\)

So, they produced a total of \(58240 + 1400 = \color{brown}\boxed{59640}\)

-2 and 4

https://math.stackexchange.com/questions/3825841/sherlock-holmes-base-number-problem

If we plot the given info in a Venn diagram, it would look like this: 

We will first solve for the number of people who only play cricket. 

Since we know the total is 55, we can write an equation: \(c + 13 + 3 + 25 = 55\), meaning \(c =14\)

Next, we know that \(b + x + 16 = 35\) and \(f + x + 28 = 53\). 

But, knowing that there are 90 students in total, we can write the equation \(b + x + f = 90\). 

Now, just solve for f. Hint: Simplify the equations and substitute 1 into another

maybe format your question properly?

If a quadratic has a repeated root, the discriminant (\(b^2 - 4ac\)) is equal to 0. 

Can you take it from here? Feel free to ask if you need more help

I think it has to do with the reflection and rotation part...

                                                                                                                                                                                                                     1 2 3

For example, with Chris's method, he counts for each of these below, which can all be reflected and or rotated to get the original grid (4 5 6) 

                                                                                                                                                                                                                     7 8 9  

1 2 3    7 4 1    9 8 7    3 6 9       

4 5 6    8 5 2    6 5 4    2 5 8

7 8 9    9 6 3    3 2 1    1 4 7

3 2 1    9 6 3    7 8 9    9 6 3

6 5 4    8 5 2    4 5 6    8 5 2

9 8 7    7 4 1    1 2 3    7 4 1

7 8 9    1 4 7    3 2 1    1 4 7

4 5 6    2 5 8    6 5 4    2 5 8

1 2 3    3 6 9    9 8 7    3 6 9

9 8 7    3 6 9    1 2 3    7 4 1

6 5 4    2 5 8    4 5 6    8 5 2

3 2 1    1 4 7    7 8 9    9 6 3

There are 4 ways to rotate the square, and 4 ways to reflect it, so i think it is \(161280 \div 16 = \color{brown}\boxed{10080}\) ways

Factor out x from the numerator of the fraction: \(x(x + 2x^3 + 3x^5 + \cdot \cdot \cdot + 1005x^{2009})\)

Now, factor out 2 from the denominator: \(2(x + 2x^3 + 3x^5 + \cdot \cdot \cdot + 1005x^{2009})\)

So we have \({x(x + 2x^3 + 3x^5 + \cdot \cdot \cdot + 1005x^{2009}) \over 2(x + 2x^3 + 3x^5 + \cdot \cdot \cdot + 1005x^{2009})} = \color{brown}\boxed{x \over 2}\)

The length of the diagonal from one corner to the opposite is just \(\sqrt{x^2 + y^2 + z^2}\)

We have \(4(x + y + z)= 64\), meaning \(x + y + z = 16\)

We also know that \(2(xy + xz + yz) = 56\).

Note that\((x+y+z)^2 = x^2+y^2+z^2 + 2xy + 2xz + 2yz = 16^2 = 256\)

But, by the magic of substitution, we already know that \(2xy + 2yz + 2xz = 56\). Subbing this in gives us \(x ^2 + y^2 + z^2 = 200\).

So, the length of the diagonal is \(\sqrt{x^2 + y^2 + z^2} = \sqrt{200} = \sqrt{100} \times \sqrt{2} = \color{brown}\boxed{10 \sqrt 2}\)

\(26 + 44 + 47 + 59 + 34 = 210\) classes were taken in total. 

If each person took 3 classes, there are \(210 \div 3 = \color{brown}\boxed{70}\) students.