Which graph represents a geometric sequence? On a coordinate plane, 5 points are plotted. The points are (1, 1),(2, 3), (3,5), (4, 7), (5, 9). On a
coordinate plane, 4 points are plotted. The points are (1, 1), (2, 2), (3, 4), (4, 8).
On a coordinate plane, 3 points are plotted. The points are (1, 2), (4,4), (9,6).
On a coordinate plane, 3 points are plotted. The points are (1, 1), (2, 4), (3,9).

B. On a  coordinate plane, 4 points are plotted. The points are (1, 1), (2, 2), (3, 4), (4, 8).

Step-by-step explanation:

Geometric sequences are those in which each successive term is equal to the previous term multiplied by a certain common ratio.

In terms of graphs, we usually look at the y-values as part of the geometric sequences.

Look at the answer choices:

- A: (1, 1), (2, 3), (3,5), (4, 7), (5, 9)

The y-coordinates are 1, 3, 5, 7, 9. Do they have a common ratio?

3 / 1 = 3

5 / 3 = 1.6666...

Clearly, the ratio isn't common, so it's not A.

- B: (1, 1), (2, 2), (3, 4), (4, 8)

The y-coordinates are 1, 2, 4, 8. Do they have a common ratio?

2 / 1 = 2

4 / 2 = 2

8 / 4 = 2

Yes, the ratio is common, so this is our answer.

- C: (1, 2), (4,4), (9,6)

The y-coordinates are 2, 4, 6. Do they have a common ratio?

4 / 2 = 2

6 / 4 = 1.5

The ratio is not the same, so eliminate C.

- D: (1, 1), (2, 4), (3,9)

The y-coordinates are 1, 4, 9. Do they have a common ratio?

4 / 1 = 4

9 / 4 = 2.25

No, the ratios aren't the same, so D isn't right.

The correct answer is thus B.

~ an aesthetics lover

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

yes

step-by-step explanation: