Mr. Martin’s math test, which is worth 100 points, has 29 problems. Each problem is worth either 5 points or 2 points. Let x be the number of questions worth 5 points and let y be the number of questions worth 2 points. x + y = 29, 5x + 2y = 100

The number of questions worth 5 points is 14 and the number of questions worth 2 points is 15. 70 points can be gained from the 5 point questions and 30 points can be gained from the 2 point questions making a total of 100 points.

Step-by-step explanation:

For this problem, we have to solve for x and y. We are given two equations.

x + y = 29

5x + 2y = 100

Let's find the value of x first. We can rearrange the first equation so it'll be easier to find the value of x using the second equation.

x + y = 29 → y = -x + 29

Now, let's plug in this equation into the second equation.

5x + 2(-x + 29) = 100

Distribute 2 to (-x + 29)

5x - 2x + 58 = 100

Combine like terms.

3x + 58 = 100

Subtract 58 from 100.

3x = 42

Divide by 3 on both sides.

x = 14

So, the value of x is 14.

Now, let's find the value of y. We can do this by plugging in our x into the first equation.

14 + y = 29

Subtract 14 from 29.

y = 15

So, the value of y is 15.

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).