Answer:
Step-by-step explanation:
You are being asked to compare the value of a growing infinite geometric series to a fixed constant. Such a series will always eventually have a sum that exceeds any given fixed constant.
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<h3>a)</h3>
Angelina will get more money from the Choice 1 method of payment. The sequence of payments is a (growing) geometric sequence, so the payments and their sum will eventually exceed the alternative.
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<h3>c)</h3>
For a first term of 1 and a common ratio of 2, the sum of n terms of the geometric series is given by ...
Sn = a1×(r^n -1)/(r -1) . . . . . . . . . . series with first term a1, common ratio r
We want to find n such that ...
Sn ≥ 1,000,000
1 × (2^n -1)/(2 -1) ≥ 1,000,000
2^n ≥ 1,000,001 . . . . add 1
n ≥ log(1,000,001)/log(2) . . . . . take the base-2 logarithm
n ≥ 19.93
The total Angelina receives from Choice 1 will exceed $1,000,000 after 20 days.
One way ti find the common denominatir is to check ti see if ine denominator is a factor to the other deniminator if it is then the deniminator can be used as the common denominator when the two deniminators are the same compare the numerators
Answer:
D. tan(-π/6)
Step-by-step explanation:
The tangent function is periodic with a period of π, so ...
tan(5π/6) = tan(5π/6 ± kπ) . . . for any integer k
For k = -1, we have ...
tan(5π/6) = tan(5π/6 -π) = tan(-π/6)
The type of graph that would allow us to quickly see how many students were treated would be a Bar graph.
