NotThatSmart

First, we must figure out what the common difference in the arithmetric sequence is. Let's let that be d. 

From the problem, we have the equation

$1/4 + 10d = -1/5 \\ 10d = -1/5 - 1/4 = -9/20 \\ d = -9/200$

Now, we can use all the information we have to find the 43rd term. We have

$-1/5 + 10d = -1/5 + 10 (-9/200) = -1/5 - 9/20 = -65/100 = -13 / 20$

So our answer is -13/20. 

Thanks! :)

Let's first focus on the first equation given.

Since we know that $(x+y)^2 = x^2+2xy+y^2$, squaring both sides on the first equation, we get that

$x^2+2xy+y^2=4$

Reaaranging the second equation a bit, we find that

$x^2+xy+y^2=5$

Wait! These two equations are quite similar to each other. Subtracting the second equation from the first equation, we get that

$x^2+2xy+y^2=4\\ x^2+1xy+y^2=5 \space \space \space \space -$

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$xy=-1$

Now, let's look at the second equation again. We know that

$x^2+xy+y^2=5\\ x^2+y^2=5-xy\\ x^2+y^2=5-(-1) = 6$

So the answer is 6...i think

Thanks! :)

We can use a series of equations to solve this problem. 

First, let's focus on what we are trying to compute, a^3+b^3. 

Now, note that

$a^3 + b^3 = (a + b) ( a^2 + b^2 - ab)$

Wait! We already have all the terms needed to solve this problem. We have

$a^3+b^3=4(6+ab - ab) = 4(6) = 24$

Thus, our answer is just 24. 

Thanks! :)

The smallest possible prime divisor a number can have is of course, 2

If we check the multiplication, we can see that 3 odd numbers can't possibly mulitply to be an even. 

So, $5^{19} * 7^{13} * 3^{31}$  is not divisble by 2. 

The next smallest is 3. 

The third term of the multiplication is $3^{31 }$

This means, no matter what, this number has to be divisible by 3, since 3 is part of the factorization, meaning it is a factor. . 

So 3 is our final answer. 

Thanks! :)

Well, let's note something about this problem really quickly. 

So, throughout the first 12 games, you have 1 win and 11 losses, which is 1/12, or about 8% win rate. 

After that, you win 1 game and lose 2 games, which is 1/3 or about 33% win rate. 

However, in order to achieve a 60% win rate, you must win more games than you lose. 

Since you ALWAYS lose more than you win, it's not possible you have a 60% win rate. 

Thanks! :)

~900 answers

Let's note that 1/17 is a repeating decimal. 

$1/17 = 0.\overline{ 0588235294117647}$

There are 16 repeating digits in the decimal. 

Dividing 4000 by 16, we get

$4000/16 = 250$

Since it is divisble, then we can conclude that the last digit of the repeating decimal is the correct answer. 

Thus, 7 is the final answer. 

Thanks! :)

We can use the slope the equation and power of a point to solve this question. We have

$[ b^2 - a^2 ] / [ b - a ] = [ (b -a) (b + a) ] / (b -a) = b + a = 2$

So the answer is 2. 

Thanks! :)