Answer:
4) v=kx^2/y^3
2= k(4^2)/(3^3)
2 =k16/27
k = 2(27/16) = 27/8
v = (27/8)(3^2)/(2^3 = (27/8)(9/8) = 243/64
v = 243/64 when x=3 and y=2
Step-by-step explanation:
here
17) 25 questions on the test
You did it right (almost, I got 21 instead of 19) but didn't finish. You need to show your discriminant is never negative.
x² + (p+1)x = 5-2p
x² + (p+1)x +(2p-5) =0
Real roots mean a positive (or at least non-negative) discriminant:
D = b² - 4ac = (p+1)² - 4(1)(2p - 5) = p² + 2p + 1 - 8p + 20
D = p² - 6p + 21
It's not totally obvious that D>0; we prove that by completing the square by noting
(p-3)² = p² - 6p + 9
so
p² - 6p = (p-3)² - 9.
D = p² - 6p + 21
D = (p-3)² - 9 + 21
D = (p-3)² + 12
Now we clearly see D>0 always because the squared term can't be negative, so D is always at least 12. We always get two distinct real roots.
1.29,31
2.42,47
3.81,69
4.50,36
5.7,16
6.26,30
7.15,23
8.28,34
