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:
False, 5 is not a factor of 54.
Draw a rectangle with diagonal 5 in. Inside this rect. are 2 acute triangles of hypotenuse 5. Note that 3^2 + 4^2 = 5^2; thus the width of the rect. is 3 and the length is 4, with the result that the hypo. is sqrt(3^2+4^2), as expected.
Answer:
A = 27π cm²
Step-by-step explanation:
Area of the shaded region = area of full circle of O - Area of the unshaded sector
= (πr²) - (θ/360*πr²)
Where,
r = 6 cm
θ = 360
Area of shaded region = (π*6²) - (90/360*π*6²)
= (36π) - (¼*π*36)
= 36π - 9π
Area of shaded region = 27π cm²