Combinatorial Enumeration. That whole class was a rollercoaster ride of mind-blowing generating functions to prove crazy things. The exam had ridiculous questions like 'count the number of cactus trees with n vertices such that etc etc etc' and you'd do three pages of terrible terrible sums and algebra. Then your final answer would be something beautiful like n/2 and you'd breath a sigh of relief and thank the math gods.
<em>To solve for a variable, you just need to isolate the variable to one side.</em>
<h3>7.</h3>
For this, just divide both sides by r and <u>your answer will be 
</u>
<h3>8.</h3>
For this, divide both sides by nR and <u>your answer will be 
</u>
<h3>9.</h3>
Firstly, multiply both sides by T: 
Next, subtract FV on both sides of the equation: 
Lastly, multiply both sides by -1, and <u>your answer will be 
</u>
<h3>10.</h3>
Firstly, multiply both sides by 1000: 
Lastly, divide both sides by tc and <u>your answer will be 
</u>
The triangle lengths would be 14√2, 7√2 and 7√6.
The sides of a 45-45-90 triangle can be listed as t, t and t√2. The legs are each 14, so the hypotenuse will be 14√2.
This will also be the longest side of the 30-60-90 triangle. Its sides can be listed as t, 2t and t√3. Since the hypotenuse is the longest side, we have:
2t=14√2
Dividing both sides by 2, we have:
2t/2 = 14√2/2
t = 7√2
The remaining side will be t√3:
t = 7√2
t√3 = 7√2(√3) = 7√6
it would take 5 people 16 hours do decorate 200 biscuits
