Kari's paper cup holds 6 fluid ounces. If one ounce is equivalent to 29.6 milliliters, how many milliliters does her paper cup hold?

Equivalent measurements: teaspoon, ounce, pint, pound

You will find a table of equivalent measurements in the back of many cookbooks:

1 tablespoon = 3 teaspoons  

      1 cup = 16 tablespoon = 8 ounces  

    1/4 cup = 4 tablespoons  

    1/3 cup = 5 tablespoons plus 1 teaspoon  

     1 pint = 2 cups = 16 ounces  

    1 quart = 4 cups (2 pints) = 32 ounces  

   1 gallon = 4 quarts

    1 pound = 16 ounces

  1/2 pound = 8 ounces

  1/4 pound = 4 ounces

Note that a pint is 16 ounces of volume, while a pound is 16 ounces of weight. The popular rhyme "A pint's a pound, the world around" can help you remember this, but keep in mind that they're not really equivalent.

Doubling a recipe can be easy (just put in two of each measurement), but what if you need to divide a recipe in half? Here are some measurements that may help:

Measurement    1/2 Measurement

1/8 teaspoon   just a dash

1 tablespoon   1 1/2 teaspoons

1/4 cup        2 tablespoons

1/3 cup        2 tablespoons + 2 teaspoons

3/4 cup        6 tablespoons

Step-by-step explanation:

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).