Someone please help me on this . I need the work too!!

Student ticket: $2

Adult ticket: $5

Step-by-step explanation:

1. Setting up the system of equations

let x be the cost for a child

let y be the cost for an adult

For the monday, that would be:

33x + 42y = 276

For the tuesday, you ahve

46x + 40y = 292

2. Solving

First, simplify each equation

33x + 42y = 276

To get all the numbers smaller divide both sides by 3

11x + 14y = 92

Do the same for the other equation:

46x + 40y = 292

divide both sides by 2

23x + 20y = 146

You want to use elimination, so take the equation 11x + 14y = 92 and multiply both sides by 10

110x + 140y = 920

Take the equation 23x + 20y = 146 and multiply both sides by 7

161x + 140y = 1022

Subtract the two equations:

110x + 140y = 920

- (161x + 140y = 1022)

-51x = -102

divide both sides by -51

x = 2

Plug this into the original equation, 11x + 14y = 92:

11 * 2 + 14y = 92

22 + 14y = 92

subtract 22 from both sides

14y = 70

divide both sides by 14

y = 5

So the cost of a student ticket is $2 and the cost of an adult ticket is $5

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).