Number 8 is 300 because 10 times 3 equal 300
Answer:
x = √(a(a+b))
Step-by-step explanation:
We can also assume a > 0 and b > 0 without loss of generality. (If a and a+b have opposite signs, the maximum angle is 180° at x=0.)
We choose to define tan(α) = -(b+a)/x and tan(β) = -a/x. Then the tangent of ∠APB is ...
tan(∠APB) = (tan(α) -tan(β))/(1 +tan(α)tan(β))
= ((-(a+b)/x) -(-a/x))/(1 +(-(a+b)/x)(-a/x))
= (-bx)/(x^2 +ab +a^2)
This will be maximized when its derivative is zero.
d(tan(∠APB))/dx = ((x^2 +ab +a^2)(-b) -(-bx)(2x))/(x^2 +ab +a^2)^2
The derivative will be zero when the numerator is zero, so we want ...
bx^2 -ab^2 -a^2b = 0
b(x^2 -(a(a+b))) = 0
This has solutions ...
b = 0
x = √(a(a+b))
The former case is the degenerate case where ∠APB is 0, and the value of x can be anything.
The latter case is the one of interest:
x = √(a(a+b)) . . . . . . the geometric mean of A and B rotated to the x-axis.
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<em>Comment on the result</em>
This result is validated by experiments using a geometry program. The location of P can be constructed in a few simple steps: Construct a semicircle through the origin and B. Find the intersection point of that semicircle with a line through A parallel to the x-axis. The distance from the origin to that intersection point is x.
Answer:
a. Expected frequency of losses per semester = 0*0.08 + 5*0.16 + 10*0.28 + 15*0.32 + 20*0.14 + 25*0.02
Expected frequency of losses per semester = 11.7 losses per semester
b. Variance = (0-11.7)²*0.08 + (5-11.7)²*016 + (10-11.7)²*0.28 + (15-11.7)²*0.32 + (20-11.7)²*0.14 + (15-11.7)²*0.02
Variance = 10.9512 + 7.1824 + 0.8092 + 3.7848 + 9.6446 + 3.5378
Variance = 35.61
c. As losses equal $60, expected losses per semester = 11.7*$60 = $702
d. Expected losses of all textbooks per semester = 250*11.7*$60 = $175,500
The area is 76 ft squared, and the perimeter is 43.1 ft. CORRECT ME IF IM WRONG BECAUSE I WAS SO BAD AT AREA/PERIMETER
