Rom

well let's take a look...

it's in the 5x5 square

it's in all 4 of the 4x4 squares

it's in all 9 of the 3x3 squares

It's in 4 of the 2x2 squares

It's in 1 of the 1x1 squares

for a total of 19 squares

since f is symmetric in m and n we lose no generality in saying that m=1 and n is a free parameter

thus

f(n) = f(1)f(n) - f(1+n) + 1001 = 2f(n) - f(n+1) + 1001

f(n) = f(n+1)-1001

f(n-1) = f(n) - 1001

f(n) = f(n-1)+1001

it makes me sad that you can't figure out the first one.

any of these will do

\(c_1 a^{m_1}b^{n_1} \cdot c_2 a^{m_2}b^{n_2} \\ \text{where } \dots \\ c_1 \cdot c_2 = -84 \\ m_1 + m_2 = 6 \\ n_1 + n_2 = 10 \\ \text{use your rainbow panda powers to choose two monomial that satisfy these 3 requirements!}\)

for the second one we know (should know) that the volume of a cube is \(s^3 \text{ where s is the length of a side.}\)

our side length is doubled giving us \(s=4x^5\)

Thus the new volume is \(\left(4x^5\right)^3 = 4^3 \cdot x^{5 \cdot 3} = 64 x^8\)

\(|A \cup B| = |A|+|B| - |A \cap B|\)

let A be brown haired students, and B be right handed students

\(|A|=\dfrac 3 4 \cdot 28 = 21 \\ |B| = \dfrac 6 7 \cdot 28 = 24 \\ |A \cup B| \leq 28 \\ 21 + 24 - |A \cap B| \leq 28 \\ 17 \leq |A \cap B|\)

so the smallest number of students that have both brown hair and are right handed is 17

3 distinct letters so 3!=6 arrangements

The smaller triangle has a hypotenuse length of \(\sqrt{4^2+6^2} = \sqrt{52}=2\sqrt{13}\)

as the triangles are similar the zip line will have the same ratio to that hypotenuse as the 140 ft leg does to the 4 ft leg.

\(\dfrac{140}{4} = 35 \cdot 2\sqrt{13} = 70\sqrt{13}\)

and thus the zip line has length \(70\sqrt{13} \approx 252 ft\)

Pick a letter, there are 26 choices, that letter is the first and last letter in the word.

We then have 2 slots left and can choose any letters to fill them.  There are 26² ways of choosing them

So there are 26x26² = 26³ =17576 words with this property

your confusion comes from seeing * and thinking ordinary multiplication.

In this problem it's not.

think of * in this case as *(a,b) = 1 + ab where a and b are mutliplied using ordinary real multiplication

for example 2*5 = 1 + (2)(5) = 11

* takes two real arguments and returns a real result so yes it is a binary operator on the real numbers

For each note just assign a student number which can be repeated among the notes.

This is a 5 digit base 10 number and thus there are 10^5 = 100000 ways to distribute the problems

so you need to find g such that that infinitely nested square root is equal to 2

I showed you somthing similar earlier.

let the nested square root be x then

\(x = \sqrt{g +x} \\ x^2 =g+x \\ x^2 - x- g = 0 \\ x = \dfrac{1\pm \sqrt{1+4g}}{2} = 2 \\ 1 \pm \sqrt{1+4g}=4 \\ \pm \sqrt{1+4g}=3 \\ \text{we need only consider the + case} \\ 1+4g = 9 \\ 4g = 8 \\ g=2\)