A textbook store sold a combined total of 294 history and sociology textbooks in a week. The number of history textbooks sold wa
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Let AB be a chord of the given circle with centre and radius 13 cm.
Then, OA = 13 cm and ab = 10 cm
From O, draw OL⊥ AB
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = ½AB = (½ × 10)cm = 5 cm
From the right △OLA, we have
OA² = OL² + AL²
==> OL² = OA² – AL²
==> [(13)² – (5)²] cm² = 144cm²
==> OL = √144cm = 12 cm
Hence, the distance of the chord from the centre is 12 cm.
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Answer:
Collin: about $401 thousand
Cameron: about $689 thousand
Step-by-step explanation:
A situation in which doubling time is constant is a situation that can be modeled by an exponential function. Here, you're given an exponential function, though you're not told what the variables mean. That function is ...
In this context, P0 is the initial salary, t is years, and d is the doubling time in years. The function gives P(t), the salary after t years. In this problem, the value of t we're concerned with is the difference between age 22 and age 65, that is, 43 years.
In Collin's case, we have ...
P0 = 55,000, t = 43, d = 15
so his salary at retirement is ...
P(43) = $55,000(2^(43/15)) ≈ $401,157.89
In Cameron's case, we have ...
P0 = 35,000, t = 43, d = 10
so his salary at retirement is ...
P(43) = $35,000(2^(43/10)) ≈ $689,440.87
___
Sometimes we like to see these equations in a form with "e" as the base of the exponential. That form is ...
If we compare this equation to the one above, we find the growth factors to be ...
2^(t/d) = e^(kt)
Factoring out the exponent of t, we find ...
(2^(1/d))^t = (e^k)^t
That is, ...
2^(1/d) = e^k . . . . . match the bases of the exponential terms
(1/d)ln(2) = k . . . . . take the natural log of both sides
So, in Collin's case, the equation for his salary growth is
k = ln(2)/15 ≈ 0.046210
P(t) = 55,000e^(0.046210t)
and in Cameron's case, ...
k = ln(2)/10 ≈ 0.069315
P(t) = 35,000e^(0.069315t)