Item 3
1 answer:
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Answer:
12%
Step-by-step explanation:
The equation for the growth is ...
f(t) = (initial value)×(growth multiplier per period)^(number of periods)
where the growth multiplier is often expressed as a percentage added to 1:
multiplier = 1+r
growth rate = r
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This equation has two unknowns:
- initial value
- growth multiplier
In order to find these, you can make use of two of the supplied data points. I like to choose the ones that are farthest apart, as they tend to average out any errors due to rounding.
Clearly, the table tells you the initial value is 210. If you don't believe, you can put the numbers in the equation to see that:
f(0) = (initial value)×(growth multiplier)^0
210 = (initial value)×1
(initial value) = 210
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Using the last data point, we get ...
f(7) = 210×(growth multiplier)^7
464 = 210×(growth multiplier)^7 . . . . . . . . . fill in table value
2.209524 = (growth multiplier)^7 . . . . . . . divide by 210
You can solve this a couple of ways. My calculator is able to take the 7th root, so I can use it to find ...
Alternatively, you can use the 1/7 power:
2.209524^(1/7) = (growth multiplier)
Another way to solve this is to use logarithms:
log(2.209524) = 7×log(growth multiplier) . . . . . take the log
log(2.209524)/7 = log(growth multiplier) . . . . . divide by 7
0.04918553 ≈ log(growth multiplier)
growth multiplier = 10^0.04918553 ≈ 1.11992 . . . . take the antilog
So, our growth multiplier is ...
1 + r ≈ 1.11992
r ≈ .11992 ≈ 12.0%
The rate of growth is about 12% in each period.
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Collapsing all of that to a single calculation:
growth rate = (464/210)^(1/(7-0)) -1 ≈ 12.0%
Part (a)
<h3>Answer: y1 and y3 are perpendicular</h3>This is because the two slopes 2 and -1/2 multiply to -1. Perpendicular slopes multiply to -1 assuming neither line is vertical or horizontal.
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Part (b)
Graph each line to see where they cross. The three points of intersection are
(0,4)
(2,-2)
(4,2)
The order of the points doesn't matter.
You could also form three systems of equations pairing up the equations, and solving each system. That way you can find the points of intersection. Graphing may be a better and faster route in my opinion. See the diagram below.
