if we have a geometric series with u1 = 2.1 and r= 1.06, what would the least value of n be such that Sn > 5543?
1 answer:
Answer:
88
Step-by-step explanation:
Write the expression for the sum in the relation you want.
Sn = u1(r^n -1)/(r -1) = 2.1(1.06^n -1)/(1.06 -1)
Sn = (2.1/0.06)(1.06^n -1) = 35(1.06^n -1)
The relation we want is ...
Sn > 5543
35(1.06^n -1) > 5543 . . . . substitute for Sn
1.06^n -1 > 5543/35 . . . . divide by 35
1.06^n > 5578/35 . . . . . . add 1
n·log(1.06) > log(5578/35) . . . take the log
n > 87.03 . . . . . . . . . . . . . . divide by the coefficient of n
The least value of n such that Sn > 5543 is 88.
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The one way to determine factors of x³ + 11x² – 3x – 33 will be .
It is the method to separate the polynomial into parts and the parts will be in multiplication. And the value of the polynomial at this point will be zero.
The steps involved in the factorization are;
1. For each pair of parentheses, we create a common factor.
2. We use x2 as a common factor for the first parenthesis.
3. We use common factor 3 for the second parenthesis.
We will find the final solution as .
Hence the one way to determine factors of x³ + 11x² – 3x – 33 will be .
To learn more about the factorization refer to the link;
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,
Step-by-step explanation:
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