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.
Answer:
Answer to fourth part is :
Angle ABC= Angle DBC
So for fifth part reason answer is ASA congruency.
Step-by-step explanation:
Here we are given that CB bisects angle ABD and angle ACD.
So we have ,
Angle ABC= Angle DBC
Answer to fourth part is :
Angle ABC= Angle DBC
Now here we have two angles and one side equal in two triangles.
So we can say that ASA congruency fits in the best here.
So for fifth part reason answer is ASA congruency.
