(4/500)m = (1/ 125)m in one second
Taking the cosine inverse, we have
cos-1(.96) = A ≈ 16.26°
This angle might also lie in the 4th quadrant ≈ (360 - 16.26)° ≈ 343.74°
(1/5)n:16=256:20n cross-multiply
(1/5)(20)n^2 = 16*256
4n^2 = 16*256 divide both sides by 4
n^2 = 4 *256 take the square roots of both sides
n = (+/-)sqrt(4)*sqrt(256)
n= (+/-) 2*16
n = (+/-)32
= 0.00243
|4x+1|=10 we have two equations
4x + 1 = 10 and 4x + 1 = -10
subtract 1 from both sides
4x = 9 4x = -11
divide both sides by 4
x = 9/4 and x = -11/4
And those are the two solutions
Let's suppose that the game started at 12 noon and lasted 3 hrs and 36 minutes.......then...it would have been over at 3:36PM
But...it started 10 minutes earlier, so it ended 10 minutes earlier.....i.e., 3:26PM
We need to find the variation constant, k, in this type of equation:
y= kx so we have
2.52 = k(8.4) divide both sides by 8.4
k = 2.52/8.4 = .3
So .....when x = 2.7....we have that :
y =(.3)2.7 = .81
9.2 x 106 = 9,200,000
No..... an x cannot be associated with two different y values
(5,7) and (5,8) violate this
