Atotal of 20 quarters and nickels add up to $4.00. how many nickels are there?

5 nickels

Step-by-step explanation:

You can setup and solve a system of equations, or you can solve by trial and error until you get the correct answer.

Here is the solution by trial and error.

If all 20 coins are quarters, the value is 20 * $0.25 = $5

That is too much value.

Let's try 16 quarters. 16 quarters are worth 16 * $0.25 = $4.

That is the correct value, but it is only with quarters, and only 16 of them.

We need fewer quarters than 16.

Try 12 quarters: 12 * $0.25 = $3.00

The number of nickels is: 20 - 12 = 8

8 nickels are worth 8 * $0.05 = $0.40

12 quarters and 8 nickels are worth $3.00 + $0.40 = $3.40

There are 20 coins, but the value is too low.

The number of quarters is between 12 and 16.

Try 14 quarters and 6 nickels:

14 * $0.25 + 6 * $0.05 = $3.50 + $0.30 = $3.80

We are closer to $4 but not there yet.

Try 15 quarters and 5 nickels.

15 * $0.25 + 5 * $0.05 = $3.75 + $0.25 = $4

The total value is $4 and there are 20 coins. This is the answer.

15 quarters and 5 nickels works.

5 nickels

its kind of blurry. can you put a clearer pic?

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).