Create truth table to determine whether or not the following statements are logically equivalent
1 answer:
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Very nice problem. You should look up the history of this question. It has a very long one having to do with mirrors made between 50 BC and 50 AD.
Step One
Find AB, BC, AC
Before beginning this problem, I'm going to arbitrarily name
c = AB
a = BC
b = AC It just makes the calculations easier.
AB = radius of the large circle - radius of the medium circle = 20.62 - 8.04 = 12.58
BC = Radii of the two smaller circles added together = 8.04 + 7.35 = 15.39
AC = Radius of the large circle - Radius of the small circle = 20.62 - 7.35 = 13.27
So
a = 15.39
b = 13.27
c = 12.58
Step Two
Find Angle A or <BAC
a^2 = b^2 + c^2 - 2*b*c * Cos(A)
15.39^2 = 12.58^2 + 13.27^2 - 2 * 12.58 * 13.27 * Cos(A)
236.852 = 158.26 + 176.1 - 333.87 * Cos(A)
236.852 = 334.4 - 333.87*Cos(A)
-97.55 = - 333.87 * Cos(A)
-97.55 / -333.87 = Cos(A)
0.2922 = cos(A)
A = cos-1(0.2922)
A = 73.01 degrees
Step Three
Just to see if you understand what was done, I'll give you the givens for finding c, and the answer and you can work through the calculations to see if your answer agrees with mine. If it doesn't PM me.
c = 12.58
b = 13.27
a = 15.39
12.58^2 = 13.27^2 + 15.39^2 - 2*13.27*15.39*Cos(C)
C = 51.42
Step Four
Find <B
Every triangle has 180 degrees so
B = 180 - <A - C
B = 180 - 51.42 - 74.01
B = <54.57.
I have found a moderator who opened the question up so that I can show you why the Cos law is the only way to do this. If the circles are very disproportionate as in this diagram, then no simple assumption can be made. The cos law is all that will work. I would have posted this earlier, but I didn't think anyone would find another method. It's ingenious but not possible for the situation below.
Step One
Find AB, BC, AC
Before beginning this problem, I'm going to arbitrarily name
c = AB
a = BC
b = AC It just makes the calculations easier.
AB = radius of the large circle - radius of the medium circle = 20.62 - 8.04 = 12.58
BC = Radii of the two smaller circles added together = 8.04 + 7.35 = 15.39
AC = Radius of the large circle - Radius of the small circle = 20.62 - 7.35 = 13.27
So
a = 15.39
b = 13.27
c = 12.58
Step Two
Find Angle A or <BAC
a^2 = b^2 + c^2 - 2*b*c * Cos(A)
15.39^2 = 12.58^2 + 13.27^2 - 2 * 12.58 * 13.27 * Cos(A)
236.852 = 158.26 + 176.1 - 333.87 * Cos(A)
236.852 = 334.4 - 333.87*Cos(A)
-97.55 = - 333.87 * Cos(A)
-97.55 / -333.87 = Cos(A)
0.2922 = cos(A)
A = cos-1(0.2922)
A = 73.01 degrees
Step Three
Just to see if you understand what was done, I'll give you the givens for finding c, and the answer and you can work through the calculations to see if your answer agrees with mine. If it doesn't PM me.
c = 12.58
b = 13.27
a = 15.39
12.58^2 = 13.27^2 + 15.39^2 - 2*13.27*15.39*Cos(C)
C = 51.42
Step Four
Find <B
Every triangle has 180 degrees so
B = 180 - <A - C
B = 180 - 51.42 - 74.01
B = <54.57.
I have found a moderator who opened the question up so that I can show you why the Cos law is the only way to do this. If the circles are very disproportionate as in this diagram, then no simple assumption can be made. The cos law is all that will work. I would have posted this earlier, but I didn't think anyone would find another method. It's ingenious but not possible for the situation below.
Answer:
Inequalities are,
y ≥ 4x + 2
y ≥ 2
Step-by-step explanation:
Solid yellow line of the graph attached passes through two points (0, -2) and (1, 2).
Let the equation of this line is,
y = mx + b
Slope of the line =
m =
m = 4
Y-intercept 'b' = -2
Equation of the line will be,
y = 4x - 2
Since shaded area is on the left side of this solid line so the inequality representing this region will be,
y ≥ 4x - 2
Another line is a solid blue line parallel to the x-axis.
Shaded region (blue) above the line will be represented by,
y ≥ 2
Therefore, the common shaded area of these inequalities will be the solution of the given inequalities.