NotThatSmart

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! :)

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

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

______________________

$xy=-1$

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! :)

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

Let's multiply the polynomials and see what we get. 

We have $(x^3)(x^2 + tx)$

Distributing in the x^3, we find that we get the expression $x^5+tx^4$

there is actually no x^2 term, meaning t can pretty much be anything. 

As long as t doesn't contain a x^-2 term, then it isn't possible there is an x^2 term. 

Thanks! :)

For this problem, we have to go throug every function and find when f(x) = x. 

First off, we have $f(x)=2x+8$. 

We have to solve $2x+8=x$, in which we get $x=-8$. However, -8 isn't in the range $1 \le x \le 2$, so it isn't a valid solution. 

Next, we have $f(x) = 13 - 5x$

Solving for $13-5x=x$, we get $x=13/6$. Indeed, 13/6 is in the range $2 < x \le 3$, so it is a valid solution. 

After, we have $f(x) = 20 - 14x$

Solving for $20-14x=x$, we get $x=4/3$, which is not a valid solution since it's not in the range $3 < x \le 4$. 

Lastly, we have $f(x)= 40 + 5x$. 

We have $40+5x=x$, in which we get $x=-10$, which is not a valid solution. 

So, we only have x = 13/6

Thanks! :)

For there to be no real solutions, the discriminant needs to be NEGATIVE

The descriminant is $b^2 - 4ac <0$  with  $b = k ;   \space a = 5   ; \space c = 8$

Plugging it all in, we have

$k^2 - 4(5)(8) <0 \\ k^2 < 160$

$k <   12.65$      

We are trying to find the largest INTEGER, which would be 12. 

Thanks! :)

$XZW$ will form a triangle $\triangle XZW$

$\angle ZXW  = 90^\circ \\ \angle XWZ = 60^\circ$

So $\angle XZW =  180^\circ - 90^\circ - 60^\circ  =   30^\circ$

Thanks! :)

Let's first take a look at the fraction 1/17. 

It has 16 repeating digits in the form of $0. \overline{0588235294117647}$

We simplfy have to find which digit the 4000th lies on. 

First, let's do 4000/16. 

We get

$4000/16=250$

So the 4000th digit is the last digit of the repeating decimal, which is 7. 

So 7 is our answer. 

Thanks! :)