Please help me answer them they are due in 15 mins!?

1. 2

2. 3

3. 6

4. 3

5. 6

6. 3

7. -6

8. -3

9. 3

10. -5

11. 12

12. -18

13. -1

I'm sorry if any of these are wrong I didn't get to check them over.

Step-by-step explanation:

I will be substituting "a number" with the variable x.

1. Four more than twice a number is 8.

2x+4 = 8

2x = 4

x = 2

2. Three more than four times a number is 15.

4x+3 = 15

4x = 12

x = 3

3. Five less than twice a number is 7.

2x-5 = 7

2x = 12

x = 6

4. One less than four times a number is 11.

4x-1 = 11

4x = 12

x = 3

5. Seven more than the quotient of a number and 2 is 10.

(x/2)+7 = 10

(x/2) = 3

x = 6

6. Six less than six times a number is 12.

6x-6 = 12

6x = 18

x = 3

7. Five less than the quotient of a number and 3 is -7.

(x/3)-5 = -7

x/3 = -2

x = -6

8. Seven more than twice a number is 1.

2x+7 = 1

2x = -6

x = -3

9. The difference between 5 times a number and 3 is 12.

5x-3 = 12

5x = 15

x = 3

10. Nine more than 3 times a number is -6.

3x+9 = -6

3x = -15

x = -5

11. Nine more than the quotient of a number and 4 is 12.

(x/4)+9 = 12

x/4 = 3

x = 12

12. Four less than the quotient of a number and 3 is -10.

(x/3)-4 = -10

x/3 = -6

x = -18

13. Nine less than six times a number is -15.

6x-9 = -15

6x = -6

x = -1

b

step-by-step explanation:

the word superfluous means unnecessary

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).