My bad, it screwed up when I first tried to put it in, had url in there for some reason :S Fixed it nowYea..why did you repear so many times?
It is right however
Good to know it's right, was confused for a sec there hahaha.
My bad, it screwed up when I first tried to put it in, had url in there for some reason :S Fixed it nowYea..why did you repear so many times?
It is right however
But they have to satisfy both equations.For Q2.
It's funny.
For the first part A and B could be 0
But for the Tan part, A can be 60 and B can be 30. Or 0
Thx man. Yea, I don't know how to post pics on BOS. Teach me that too .hscishard - pl type out question at leat for benefit of readers
If I knew how to post figures, diagrams, graphs . . . I wouldn't have posted such a long-winded solution.Thx man. Yea, I don't know how to post pics on BOS. Teach me that too .
Wth..I got 34.54 and 35.46.I don't want this thread to die
New Question:
A piece of wire is 70 cm long. One section is bent into an equilateral triangle and the other into a rectangle that is three times as long as it is wide. Find the lengths of the two pieces of wire if the sum of the areas of the two shapes is a minimum.
I did it in a similar way. Except I let x = perimeter of triangle.
Then make it into a general quadratic expression and find the minimum value of the parabola.
What the hell. Then where is my flaw?A = 1/2 ab sin C
A = x2/2 * √3/2
A = x2√3/4