A concrete driveway is 5 feet longer than twice its width. It has a brick walkway 3 feet wide on the front and side of the driveway. The area of the brick walkway is 213 square feet. Find the dimensions of the driveway.

The dimensions of the driveway are 21 ft x 47 ft, which is option "D" in your list of possible answers

Step-by-step explanation:

Notice that the length "L" of the driveway can be written in terms of its width as:

L = 2 w + 5

based on the given information.

Please see the attached diagram to understand the geometry of the problem, and what the area of the brick walkway involves (depicted in brick color) as the addition of the area of two rectangles:

1) one of length L= 2 w + 5 (w being the width of the driveway - depicted in grey) times 3 feet (the width of the brick walkway

and the other one:

2) w + 3 times 3

Therefore the total area of the brick walkway is given by the formula:

(2 w + 5) * 3 +(w + 3) * 3 = 213 square feet

Now we can solve for "w" in the above equation:

6 w + 15 + 3 w + 9 = 213

9 w + 24 = 213

9 w = 213 - 24

9 w = 189

w = 189/9

w = 21 feet

We can now solve for the L of the driveway, using the first formula we created:

L = 2 w + 5

L = 2 (21) + 5

L = 42 + 5

L = 47 feet

Therefore the driveway's width is w = 21 feet

and its length is L = 47 feet

nada

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)

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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).