tertre

Use complementary counting...

I think I answered this problem few weeks ago...Here: https://web2.0calc.com/questions/help_22327#r1

Hint: Partial Fraction Decomposition.

Right, gives 10*11=110/2=55-1=54. 

It's an equilateral triangle.

The distance formula is \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\)

1. As you can use Heron's formula for the first question, we have an equilateral triangle since a=b=c=1.

The formula to find the area of an equilateral triangle is \(\frac{\sqrt{3}}{4}s^2\)\(\), so we have  \(\frac{\sqrt{3}}{4}1^2=\frac{\sqrt{3}}{4}\approx\boxed{0.433}.\)

2. Using Heron's Formula, we have: \(\sqrt{(50)(50-49)(50-33)(50-18)}=40\sqrt{17}\approx\boxed{164.92}.\)

1 (a): https://www.desmos.com/calculator/5g7mmiuq61

1 (b): https://www.desmos.com/calculator/bh7n5oj7zi

What do you see ?

I think the guest means:

Given that \(-3 \le 7x + 2y \le 3\) and \(-4 \le y - x \le 4\), what is the maximum possible value of  \(x+y\) ?

Please write out your equations not just a graph.  Thanks a ton.

An approximation could be \(\frac{22}{7}\) or \(\frac{355}{113}.\)

The slope of the line can be found by: \(\frac{y_2-y_1}{x_2-x_1}\) , so we have \(\frac{-2-5}{4-3}=\boxed{-7}.\)

\(\sqrt{37+20\sqrt{3}}=a+b\sqrt{3}\)- Original Equation; trying to solve for \(a,b\).

Square both sides, to get \({37+20\sqrt{3}}=a^2+2ab\sqrt{3}+3b^2.\)  This is just the expansion of \((a+b)^2\) !

Now, we use a bit of matching, on trying to match the variables to the numbers. We see  \(20\sqrt{3}\) and \(2ab\sqrt{3}\) , so we have \(2ab\sqrt{3}=20\sqrt{3}\) , and \(ab=10.\)

Also, we get \(a^2+3b^2=37\) by matching the variables to the numbers, again.  

After trying a few times, we get \(a=5, b=2\) , and that sure works !

Check: \((5)(2)=10\) and \(5^2+3(2)^2=25+12=37.\)

Thus, the answer is \(5,2\) .