NotThatSmart

We can probably write some equations with the information we have. 

We have:

\(a_n = a_{n - 1} + a_{n - 2}\)  Subtracting both sides by a_n-1, we have

\(a_{n-2}=a_n-a_{n-1}\)

Now, let's plug in numbers we are given in the question. Plugging in 11 gets us \(a_9=4-1=3\)

Plugging in 10 gets us \(a_8=1-3=-2\)

Ok, let's see what happens when we plug in 9. 

We get \(a_7 = a_9 - a_8\). We already know these two values!

We have \(a_7=3-(-2)=5\)

Now, we plug in 8. We get \(a_6=-2-5=-7\)

So -7 is our final answer.

Thanks! :)

First, let's find every possible interval with this equation. 

We have \(c<-5,\:\:-5\le \:c<0,\:\:0\le \:c<4,\:\:c\ge \:4\)

Now, we solve the equation for each inequality. 

Case 1 - \(c<-5\)

\(-c-5-3c = 10 - 2c+8+6c\\ -8c - 23 = 0\\ c = -\dfrac{23}8\)

This root is greater than -5, so it is invalid. 

Case 2 - \(-5 \leq c < 0\)

\(c + 5 - 3c = 10 - 2c + 8 +6c\\ -6c - 13 = 0\\ c = -\dfrac{13}6\)

This works!

Case 3 - \(0 \leq c < 4\)

\(c + 5 - 3c = 10 - 2c + 8 -6c \\ 6c = 13\\ c = \dfrac{13}6\)

This works!

Case 4 - \(c>4\)

\(c + 5 - 3c = 10 + 2c - 8 - 6c\\ 2c + 3 = 0\\ c = -\dfrac32\)

This is invalid. 

Now, we have \(c = -\frac{13}6, \frac{13}{6}\)

Thanks! :)

With the first equation, let's square both sides. We get

\(a^2 + 2ab + b^2 = 16 \\ a^2 + b^2 = 16 - 2ab \)

Now, we have a second equation of 

\(a^2 + b^2 = 6 + 2ab \)

Now, we simplfy subtract the first equation from the second one, we get 

\(0 = -10 + 4ab \\ 10 = 4ab \\ ab = 10 / 4 = 5/2 \\ a^3 + b^3 = (a + b) ( a^2 + b^2 - ab) \\ a^3 + b^3 = (4) ( 6 + 2ab - ab) \\ a^3 + b^3 = (4) ( 6 - ab) \\ a^3 + b^3 = (4) ( 6 - 5/2) \\ a^3 + b^3 = (4) ( 7/2) \\ a^3 + b^3 = 14\)

14 is our answer. 

Thanks! :)

We can easily just list the values we have 

\(\frac{1}{3}, \frac{2}{3}, \frac{4}{3},\frac{5}{3}, \frac{7}{3}, \frac{8}{3}, \frac{10}{3}, \frac{11}{3}, \frac{13}{3}, \frac{14}{3}\)

The numbers we excluded like 3/3 or 6/3 are not in simplest form, so we ignore them. 

Adding all these fractions up, we get \(75/3=25\)

So 25 is our answer. 

Thanks! :)

Well, there is actually only 1 possible. 

2-2-1 wis the only one that works. 

Others wouldn't satisfy the triangle inequality theorems. 

Thanks! :)

The smallest possible perimeter is 3. 

Let's say that b is 1. 

We would have

\(a < 1 \\ c > 1\)

where 

\( a + c = 2 \\ a > .50 \\ c < 1.5 \)

An example would be

\(a = .75\\ b = 1\\ c =1.25\)

Thanks! :)

We cam first find the probability we get each number. We have degree numbers, so let's use that to help us. 

Since 1 has a degree measure of 120 degrees, the probability we get 1 is \(120/360 = 1/3\)

Since 2 has a degree measure of 60 degrees, the probability we get 2 is \(60/360=1/6\)

3 has the same degree measure as 1, so we have \(1/3\)

4 has the same degree measure as 2, so we have \(1/6\)

This means after 6 spins, we have an expected value that

- two 1s

- two 3s

- one 4

- one 2

\(3+3-1-1-1+2 = 5\)

Now, 1000/6 = 166.67, so let's first do

\(166*5=830\)

We have 4 throws remaining. \(2/3 * 5 = 10/3\)

\(833.33 \approx 833\)

Not sure if this is correct, but it should be close.

Thanks! :)

Ok, let's first let the intersection of AD and FB be Z. 

We know AB = 8. 

Lets let FZ = x

From the problem, we can tell that triangles ZFD and ZAB are congruent triangles. From this, we know FB = 9. 

ZB = FB - FZ = 9 - x

Now, let's use the awesome pythaogrean theorem. 

\(ZB^2 = FZ^2 + AB^2 \\ (9 -x)^2 = x^2 + 8^2 \\ x^2 - 18x + 81 = x^2 + 64 \\ 81 - 64 = 18x \\ 17 = 18x \\ x = 17/18\)

Now, we can find the non-shaded area. 

\(4 * [ ABZ ] = 4 (1/2) (AB)(AZ) = 4 (1/2) (8)(17/18) = 272/18 = 136/9\)

Now, the shaded area is \(8*9 - 136 / 9 = 72 - 136/9 = 512 / 9\)

Thanks! :)

First, let's write a quadratic equation to find the two roots. 

We have \(x^{2}+9x-20=0\)

Notice that \(r^2 + s^2 = (r+s)^2-2rs\). 

Now, let's take a break from this. If we had roots r and s, we would have

\((x-r)(x-s)\\ x^2-sx-rx+rs\\ x^2-(r+s)x+rs\)

Using the equation from above, we find that, r+s is equal to -9 and rs is equal to -20. 

Plugging that in to \( (r+s)^2-2rs\), we get\((-9)^2-2(-20) = 81+40 = 121\)

So 121 is our answer!

Thanks! :)\(\)