Unit: Surface Area Homework 5
Name Stechnic Nieke
Date a
Pd 7
APPLYING SURFACE AREA OF CYLINDERS
1. Whitney buys a coffee in a cylindrical cup with a cardboard sleeve so the cup isn't too hot to
hold. If the diameter of the sleeve is 5 inches and the height of the sleeve is 4 inches, what is
the surface area of the cardboard sleeve?
2 Trh
dis
2.3.14.25.4
h=4
6=2.5
62.8
2. Melvin is pushing a tire during a workout with the dimensions shown. How many square
inches of space has the tire covered after Melvin pushes the tire 5 times around?
2 Arh+
2Trz
29 in.
Bin.
3. Each glass mug below has a radius of 2 inches and a height of 3.75 inches. If the mugs are
sold in sets of 4, find the total sur, ace area of the mugs, not including
the handles.
4. The can of peas has a radius of 3.8 centimeters and a height of 10 centimeters. The can is
made of aluminum and has a label made of paper.
a. Find the total surface area of the aluminum can.
peas
b. Find the total surface area of the label.
BIOS
5. Claude is wrapping cylindrical bug spray containers in tissue paper to put them inside a box.
Each container has a height of 12.7 centimeters and a radius of 3 centimeters.
If he is wrapping a total of 5 bug spray containers, how many square
centimeters of tissue paper will he need?
BUSS
OManeuvering the Middle LLC, 2017
I need help on homework please!!

b. x- 48 =120

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)

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

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).

c. 1,233 miles

ending amount = beginning amount / (2^n)

where "n" = number of half-lives (or elapsed time / half-life)

ending amount = 50 mg / (2 ^10 / 5.3)

ending amount = 50 mg / (2 ^ 1.8867924528)

ending amount = 50 m / 3.6981210597

ending amount = 13.5203794557 mg

or about 13.5 mg remain

source: