URGENT: Amy and Lucy were doing a craft involving yarn. Amy cut a piece of red yarn that was 4 1/3 inches long, and Lucy cut a p
iece of yellow yarn that was 2 3/5 inches long. How much longer was Amy's yarn than Lucy's? Write your answer as a mixed number, (not improper fraction,) in simplest form.
Please show steps, thank you. C:
Please show steps, thank you. C:
2 answers:
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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)
Answer:
y = ½x - 14
Step-by-step explanation:
Given the linear equation, y = 3x - 4, where the <u>slope</u>, m = 3, and the <u>y-intercept</u> is (0, -4):
The slope of a linear equation represents the steepness of the line's graph. The higher the value of the slope, the steeper the line. Hence, the slope of the other line must be less than three, but is greater than zero: 0 < <em>m</em> < 3. (a negative slope will show a <em>declining</em> line).
Next, the vertical translation of the line involves changing the value of the parent graph's y-intercept. Since the prompt states that the equation must represent a downward vertical shift of 10 units, then the y-intercept of the other line must be (0, -14).
The linear equation that I have chosen that meets the requirements of the given prompt is: y = ½x - 14. <em>You're more than welcome to choose a different slope</em>, as long as it is less than 3, but is greater than 0 (must be a positive slope).
Attached is a graph of both equations, to demonstrate that the other equation represents a graph with a steeper slope than the original graph.
