Question 3 of According to the Income section shown below from the 1040EZ form, if
$3125 were entered on which line would the taxpayer be ineligible to use the
form?
1
Wages, salaries, and tips. This should be shown in box 1 of your Form(s) W-2.
Attach your Fomm(s) W-2
Income
Attach
Form(s) W-2
here
Endose, but do
not
attach any
payment.
2
3
Taxable interest. If the total is over $1500. you cannot use Form 1040EZ
Unemployment compensation in excess of $2,400 per recipient and Alaska Permanent
Fund dividends (see page 11).
3
4
4
You may benet
from tling Form
1080 OR 1940
See Before You
Begion page 4
Add lines 1, 2 and 3. This is your adjusted gross income
If someone can claim you (or your spouse if a joint retum) as a dependent, check
the applicable boxies) below and enter the amount from the workshoct on back.
Spoase
If no one can claim you (or your spouse if a joint return, cater 59,350 if singles
$18.700 if married filing jointly. See back for explanation.
Subtract line 5 from line 4 If line 5 is larger than linc 4. eater.
This is your taxable income.
You
O A. Line 3
O B. Line 2
O C. Line 4
O D. Line 1

the answer would be c:   ax + b = ax + c

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).