23) #'s of positive integral divisors of 6^5 * 4^3 * 2^1.
i get this down to 3^5 * 4^3 * 2^6, increased all exponents by 1 and multiply to get 168, but answer key said 78.
*(60) the area of 14x^2 + 16y^2 = 224 is ____ (area of ellipse? using calculus?)
(61) 555 * 6/37 = _________
(63) 431/4 = _______ (both #'s in base 5, and answer)
23) it is 3^5 * 4^3 * 2^6, but the 4 must be factored down to 2^6, making it 3^5 * 2^12
(5+1)(12+1) = 6 * 13 = 78
60) area of ellipse is (pi * a * b)
a is sqrt(16) = 4
b is sqrt(14) = ~3.7 or 3.8
4*3.8*pi = 15*pi = 47
61) 37 * 3 = 111. So first divide 555 by 37. 3*5 = 15.
now multiply 555/37 = 15 by 6 and get 90
63) I would convert 431_5 to 116_10 and divide to get 29_10 and convert to 104_5, but there are probably better ways to do this
23) This one is a trick. Four is not a prime number. Its actually 2^6 in disguise! You actually have 2^12*3^5.
12*6 = 78.
60) Whenever you get an ellipse in the form: ax^2 +by^2 = ab, the area will always be sqrt(ab)*pi.
In this case, 14*16 = 224, so we can use the formula. So... sqrt(224) = around 15pi.
61) Remember that 111/3 = 37 or 37*3 = 111
So... you have 37*3*5 * (2*3/37) which equals 90.
On the other one... I'm stumped. I'm good with multiplying bases, but not dividing...
- Zack -
This is how I'd do 431/4:
4/4=1
31base5=16,16/4=4
therefore, 431/4=104
Best Wishes
