MathProblemSolver101

I made a mistake with the $\frac{18}{xyz}$, though I am scared to edit the latex : Do you think you can manipulate $x^2 y^2 + y^2 z^2 + x^2 z^2$ into something you already know? The answer follows from there.

$(x+y+z)^2 - 2(xy + xz + yz) = x^2 + y^2 + z^2$

$(x+y+z)^2 - 2(-7) = 18$

$(x+y+z)^2 = 4$

$x+y+z = 2$

$x^2 y^2 + 2 x^2 y z + x^2 z^2 + 2 x y^2 z + 2 x y z^2 + y^2 z^2 = 49$

$x^2 y^2 + x^2 z^2 + y^2 z^2 + 2xyz(x + y + z) = 49$

$\frac{xyz^2}{z^2} + \frac{xyz^2}{y^2} + \frac{xyz^2}{x^2} + 4xyz = 49$

$xyz \left(\frac{xyz}{z^2} + \frac{xyz}{y^2} + \frac{xyz}{x^2} \right) + 4xyz = 49$

$xyz \left(\frac{xyz(x^2 y^2 + y^2 z^2 +x^2 z^2)}{xyz^2} \right) + 4xyz = 49$

$xyz \left(\frac{18}{xyz} \right) + 4xyz = 49$

$18 + 4xyz = 49$

$4xyz = 31$

$xyz = \boxed{\frac{31}{4}}$

Minimum: You want the least numbers possible, and for those numbers you want the least of your set, namely 0 and 1. 

Maximum: You want the most numbers possible, and there isn't an upper bound, so 0 + 1 + 2 + 3 + 4 + 5 = 15 is your max.

Thus, [1, 15]

Verify:

0+1

0+2

0+3

0+4

0+5

1+5

2+5

3+5

4+5

1+2+3+4

1+2+3+5

1+2+4+5

1+3+4+5

2+3+4+5

1+2+3+4+5

Review your class transcript. This should give you a hint on how to start. Also look up the definitions and formulas relating to terminal points, and then with those you can find your answer.

$\lceil{\sqrt{300}}\rceil = \lceil{\sqrt{3 \cdot 100}}\rceil = \lceil{10 \sqrt{3}}\rceil \approx \lceil{10 \cdot 1.73}\rceil = \lceil{17.3}\rceil = \boxed{18}$

Hint: WLOG, define side lengths for the triangles given ratios.

Hint 2: Construct a line from C to B.

Hint 3: Define values for those areas.

I have done all of these, but I only have 3 equations but 4 variables. Any thoughts?

Yea... wonder who made this problem!

$\frac{j+k}{jk} = \frac{2}{3}$

$3j + 3k = 2jk$

$3j - 2jk + 3k = 0.$

Here is the graph: https://www.desmos.com/calculator/u6uyo6fqj3

Consider asymptotes and reflections. LMK if you have any more questions.