Simon makes 30 cakes gives 1/5 of the cakes to sali. he gives 10% of the 30 cakes to jane what fraction of the fake cakes does he have left

Since Simon made 30 cakes, the fraction of the cakes that Simon have left is cakes.

Given the following data:

To find what fraction of the cakes Simon have left:

In order to solve this, we would find the fraction of cake that was given to both Sali and Jane.

For Sali:

Sali = 6 cakes.

For Jane:

Jane = 3 cakes.

Now, we would determine the number left from the cake:

Remaining = 21 cakes

As a fraction:

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Total number of cakes made by Simon = 30 cakes.

Number of cakes gave to Sali = 1/5 Of total cakes = 1/5 * 30 = 6 cakes.

Number of the cakes gave to Jane = 10% of 30 cakes

10% can be written as 10/100 in fracions and in decimals it would be 0.10.

Therefore, 10% of 30 cakes = 0.10 times 30 = 3 cakes.

Total number of cakes left = Total cakes made - Cakes gave to Sali - Cakes gave to Jane = 30-6-3

Therefore, Total number of cakes left = 21.

21 cakes left our of 30 cakes.

21 out of 30 could be written in fracion form as 21/30.

We can reduce this fracion in simplest form by dividing top and bottom by 3, we get

7/10.

Therefore, 7/10 fraction of the cakes does he have left.

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).

d

step-by-step explanation: