8.EE.8 Here is the graph for one of the equations in a system of two equations. The solution to the system is (6, 2). Select all the equations that could be the other equation in the system. I need this fast! times test!

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

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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).

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.

step-by-step explanation:

ratios represent a pattern or relationship that can be scaled.

equivalent ratios are pairs of ratios that represent the same relationship between quantities, and can be generated by multiplying or dividing both quantities in the ratio by the same factor.

in this lesson, students will use double number lines and ratio reasoning to solve ratio problems. by partitioning double number lines into proportionate intervals, students will build an understanding of equivalent ratio relationships as those that are generated by multiplying or dividing two related quantities by the same number. this lesson will build upon students' prior understanding of multiplicative comparisons, number line reasoning and equivalent fractions (while highlighting the difference between equivalent fractions and equivalent ratios); and will build toward future work in this unit on building ratio tables, and work in future grades in proportionality, scaling, and expressing and graphing linear functions.

step-by-step explanation:

well if you round, things become much more simple. round 5.95 to 6, 4.25 to 4, and 1.05 to 1. now, add 6+4+1 and that would equal 11. so the items together would equal to about $11.