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! :)