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When y=2 and y=5
1. 2y-1 and (3y-5+y or 4y-5)
when y=2 ; 2(2)-1 = 3 and 4(2)-5=3
when y=5 ; 2(5)-1 = 9 and 4(5)-5=15
----nonequivalent-----
2.5y+4 and (7y+4-2y or 5y+4)
so you don't have to place any value in because 5y+4 and 7y+4-2y are equal,
whatever you place any value in, it will be all the same then
-----equivalent------
and no need to find more
I would think that all but one point would be on the line. One way to approach this problem is to find the equation of the line based upon any two points chosen at random, and then determine whether or not the other points satisfy this equation. Next time, would you please enclose the coordinates of each point inside parentheses: (2.5,14), (2.25,12), and so on, to avoid confusion.
14-12
slope of line thru 1st 2 points is m = ---------------- = 2/0.25 = 8
2.50-2.25
What is the eqn of the line: y = mx + b becomes
14 = (8)(2.5) + b; find b:
14-20 = b = -6. Then, y = 8x - 6.
Now determine whether (12,1.25) lies on this line.
Is 1.25 = 8(12) - 6? Is 1.25 = 90? No. So, unless I've made arithmetic mistakes, (1.25, 5) does not lie on the line thru (2.5,14) and (2.25,12).
Why not work this problem out yourself using my approach as a guide?
Answer:
h = 
- r
Step-by-step explanation:
The question requires you to make h the subject of the formula.
S = 2πrh + 2πr²
subtract 2πr² on both sides.
S - 2πr² = 2πrh - 2πr² - 2πr²
S - 2πr² = 2πrh
Dividing both sides by 2πr
(S - 2πr²)/2πr = 2πrh/ 2πr
h = - r
The answer would be A. There is a phase shift to the left. Because you are adding pi/6.
