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?
By solving for x you get x=6
The last option, the y-intercept.
If tan0=-3/4 and 0 is in quadrant IV, cos20= (33/25, -17/25, 32/25, 7/25, 24/25?) and tan20= (24/7, -24/7, 7/25, -7/25, 13/7, -1
Oksana_A [137]
Answer:
- cos(2θ) = 7/25
- tan(2θ) = -24/7
Step-by-step explanation:
Sometimes, it is easiest to let a calculator do the work. (See below)
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The magnitude of the tangent is less than 1, so the reference angle will be less than 45°. Then double the angle will be less than 90°, so will remain in the 4th quadrant, where the cosine is positive and the tangent is negative.
You can also use the identities ...
cos(2θ) = (1 -tan(θ)²)/(1 +tan(θ)²)
cos(2θ) = (1 -(-3/4)²)/(1 +(-3/4)²) = ((16-9)/16)/((16+9)/16)
cos(2θ) = 7/25
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tan(2θ) = 2tan(θ)/(1 -tan(θ)²) = 2(-3/4)/((16-9/16) = (-6/4)(16/7)
tan(2θ) = -24/7
