Answer: g is a geometric sequence.
Step-by-step explanation:
Each year since then, her account accumulated interest amounting to
15% of the balance in the previous year.
15% = 15/100 = 0.15
Since the balance is increasing each year, then the constant factor by which it is increasing is 1 + 0.15 = 1.15.
This factor of 1.15 is a common ratio. This means that the sequence is a geometric sequence since it is increasing by a common ratio.
I hope this helps you
at the 30-60-90 triangle has rules
30=>k
60=>k(square root of) 3
90=>2k
so in this triangle 60=> see 9
k(square root of) 3=9
k=9/(square root of) 3
2k=18/(square root of) 3
at the 45-45-90 triangle rules are
45=>2k
45=>2k
90=>[2k/(square root of 3)] (square root of) 2
x=[18/(square root of 3)].(square root of 2)
x=18(square root of 2)/square root of 3
Step-by-step explanation:
The Taylor series expansion is:
Tₙ(x) = ∑ f⁽ⁿ⁾(a) (x − a)ⁿ / n!
f(x) = 1/x, a = 4, and n = 3.
First, find the derivatives.
f⁽⁰⁾(4) = 1/4
f⁽¹⁾(4) = -1/(4)² = -1/16
f⁽²⁾(4) = 2/(4)³ = 1/32
f⁽³⁾(4) = -6/(4)⁴ = -3/128
Therefore:
T₃(x) = 1/4 (x − 4)⁰ / 0! − 1/16 (x − 4)¹ / 1! + 1/32 (x − 4)² / 2! − 3/128 (x − 4)³ / 3!
T₃(x) = 1/4 − 1/16 (x − 4) + 1/64 (x − 4)² − 1/256 (x − 4)³
f(x) = 1/x has a vertical asymptote at x=0 and a horizontal asymptote at y=0. So we can eliminate the top left option. That leaves the other three options, where f(x) is the blue line.
Now we have to determine which green line is T₃(x). The simplest way is to notice that f(x) and T₃(x) intersect at x=4 (which makes sense, since T₃(x) is the Taylor series centered at x=4).
The bottom right graph is the only correct option.
A=(2x+1)/B
C=(5x-14)/B
AC=(2x+1)/B * (5x-14)/B
AC=(10x² - 23x - 14)/BE
