NotThatSmart

mmm...not sure about my solution, but here it is. 

First, let's combine all the like terms. Isolating 115, we get

\(-8x+xy=115\)

Now factoring out x from the left side, and isolating x, we find

\(x\left(y-8\right)=115\\ x=\frac{115}{y-8}\)

Now, I guess we want integers, so let's just see what y can be to make x an integer. 

115 only has 2 factors other and 1 and itself: 5 and 23. 

So x + y has 2 cases. 

First, y could be 13, and x is 23. 

The other option is for y to be 31, and x is 5. 

So we have \(13+23=31+5=36\)

So the answer is 36. 

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! :)

mmm...let's see....

We simply have

\(slope PQ = -1 / tan [ (arctan (4) + arctan (5) ) / 2 ] = [19 - \sqrt{ 442} ] / 9 ≈ -0.225 \)

So I think the answer is -0.225. 

https://web2.0calc.com/questions/help-with-coordinates_36

Thanks! :)

It's not as hard as it might seem!

We just simply have

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

So 2 is the answer. 

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 answer is \(4\)

let me elaborate on what BRAINBOLT said.

See, if D lies on the angle bisector of BAC, and angle BAC is 60 degrees, then angle CAD cannot be 45. 

This is because since D lies on the bisector, CAD would therefore be half of BAC, which is 30. 

So, this problem is invalid, 

Thanks! :)

Let's make some observations. 

Starting with \(2!\), every other term after that is even, since it contains 2. 

Since we know that \(even + even = even\), then we have \(2! + 3! + \dots + 100! = even\)

Thus, the sequence \(2! + 3! + \dots + 100! \equiv 0 (\mod 2)\)

However, we forgot about 1 term. Since we have 1! = 1, and 1 is odd, and we know \(odd+even=odd\), then we have

\(1! + 2! + 3! + \dots + 100! \equiv 1(\mod 2)\)

Thus, the answer is 1. 

Thanks! :)

Let's focus on the two smallest numbers first. 15 and 24. 

Since N leaves a remainder, we can elmiinate the factors.

Thus, N cannot be \(3, 5, 2, 4, 6, 8, 12\)

This leaves 7, 9, 10, 11, 13, and 14.   Now, let's test out all the other numbers

15/7 = 2 R 1      24/7 = 3 R 3      so it isn't 7     

15/9 = 1 R 6      24/9 = 2 R 6      so 9 looks hopeful, we'll try another   

1457/9 = 161 R 8                        so 9 is eliminated   

15/10 = 1 R 5     24/10 = 2 R 4    so it isn't 10   

15/11 = 1 R 4     24/11 = 2 R 2     so it isn't 11   

15/13 = 1 R 2     24/13 = 1 R 11   so it isn't 13   

15/14 = 1 R 1     24/14 = 1 R 10   so it isn't 14   

mmm...I don't think there is a solution

Thanks! :)