The 7th graders at Lincoln Avenue School were helping to renovate a playground for the kindergartners at a nearby elementary school. City regulations require that the sand underneath the swings be at least 15 inches deep. The sand under both swing sets was only 12 inches deep when they started. The rectangular area under the small swing set measures 9 feet by 12 feet and required 40 bags of sand to increase the depth by 3 inches. How many bags of sand will the students need to cover the rectangular area under the large swing set if it is 1.5 times as long and 1.5 times as wide as the area under the small swing set? Show all of your work. *

The students will need 90 bags of sand to cover the rectangular area under the large swing set.

Step-by-step explanation: First of all the area of the small and large swing sets is given, all measured in feet. The sand under both swing sets was supposed to be 15 inches deep, but was found to be only 12 inches deep. The rectangular area under the small swing set was measured as 9 feet by 12 feet which gives an area measurement of 108 feet squared. In order to fill up this area by an extra depth of 3 inches, the students had to use up 40 bags of sand.

They would need to also fill up the rectangular area under the large swing set by a depth of 3 inches. However, the dimensions are different because the area under the large swing set has its length 1.5 times as long as that of the small swing set and same thing applies to its width, which is also 1.5 times that of the small swing set. This means the dimensions of the large swing set are derived as follows;

Large swing length = 9 * 1.5

Large swing length = 13.5

Large swing width = 12 * 1.5

Large swing width = 18

If the length and width of the rectangular area under the large swing set are now given as L = 13.5 and W = 18, then the area becomes;

Area = L * W

Area = 13.5 * 18

Area = 243 feet squared

We can now apply the ratio of both areas of the different swing sets to determine the correct number of bags of sand needed for the area under the large swing set.

The ratio of both areas is derived as

Ratio = 108 : 243

Ratio = 4 : 9

Let x be the number of bags of sand needed for the area under the large swing set. You now have;

4 : 9 = 40 : x

This can be properly expressed mathematically as,

4/9 = 40/x

By cross multiplication you now have,

4x = 9 * 40

4x = 360

Divide both sides of the equation by 4

x = 90

Therefore, the students will need 90 bags of sand cover the rectangular area under the large swing set.

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse

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).