X = larger number, y = smaller number
x + y = 75
x = 3y + 11
3y + 11 + y = 75
4y + 11 = 75
4y = 75 - 11
4y = 64
y = 64/4
y = 16 <=== ur smaller number
x = 3y + 11
x = 3(16) + 11
x = 48 + 11
x = 59...ur larger number
Answer:
10 fifths
Step-by-step explanation:
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.
__
<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.
__
<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:
a) x=7
b) x=4
c) x= 1/9
I hope this helps! Do you need more help?
