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3 years ago

Find the 9th term of the geometric sequence 5, -25, 125, ...

Mathematics

1 answer:

arlik [135]3 years ago

6 0

Answer:

Step-by-step explanation:

a = 5

r = -5

n = 9

t_9 = a r^(n - 1)

t_9 = (-1)^n*5 (-5) ^ 8

t_9 = (-1)^9 * 5 * (-5) ^8

t_9 = -1 ( 1953125)

t_9 = - 1953125

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3 years ago

Read 2 more answers

jolli1 [7]

Answer:

Part 1) $r=3.5\%$

Part 2) $t=6\ years$

Part 3) $21.9\ ft$

Step-by-step explanation:

Part 1) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

Let

x -----> amount of money to be invested

2x ---> Final Investment Value

we have

$t=20\ years\\ P=\$x\\A=\$2x\\r=?$

substitute in the formula above

$2x=x(e)^{20r}$

$2=(e)^{20r}$

Apply ln both sides

$ln(2)=ln[(e)^{20r}]$

$ln(2)=(20r)ln(e)$

$ln(2)=(20r)$

$0.693=(20r)$

$r=0.693/20$

$r=0.035$

$r=3.5\%$

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

Let

x -----> amount of money to be invested

2x ---> Final Investment Value

we have

$t=?\ years\\ P=\$x\\A=\$2x\\r=0.126$

substitute in the formula above

$2x=x(e)^{0.126t}$

$2=(e)^{0.126t}$

Apply ln both sides

$ln(2)=ln[(e)^{0.126t}]$

$ln(2)=(0.126t)ln(e)$

$ln(2)=(0.126t)$

$0.693=(0.126t)$

$t=0.693/0.126$

$t=5.5=6\ years$

Part 3) we have

$f(x)=18.4+2.5ln(x+1)$

where

f(x) -----> is the long jump record in feet

x -----> the number of years since the school was opened

For x=3 years

substitute

$f(3)=18.4+2.5ln(3+1)$

$f(3)=18.4+2.5ln(4)$

$f(3)=18.4+2.5(1.386)$

$f(3)=21.9\ ft$

4 0

3 years ago