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3) 4y ≥ -16
2x ≤ 4
x-2y ≥ 6

4) 2x+3y ≥ -7
x-3y ≤ 28
x+6y ≤ -26

Solve the system of linear inequalities

see below

Step-by-step explanation:

3) 4y ≥ -16

2x ≤ 4

x-2y ≥ 6

First inequality

Divide by 4

4y/4 ≥ -16/4   y   ≥ -4

second inequality

Divide by 2

2x ≤ 4     2x/2  ≤ 4 /2   x  ≤ 2

Third inequality

Subtract x from each side

x-2y ≥ 6    x-2y -x ≥ -x +6   -2y  ≥ -x +6

Then divide by -2 remembering to flip the inequality

-2y/-2 ≤ (-x+6)/-2     y  ≤1/2x -3

Graph is below

4) 2x+3y ≥ -7

Putting each in line form

3y ≥ -7  -2x

y≥ -2x/3 -7/3

Second

x-3y ≤ 28

-3y≥ -x+28

Divide and flip since it is a negative

y  ≤ 1/3x -28/3

Third

x+6y ≤ -26

6y ≤ -x-26

y ≤ -1/6x -26/6

3. The solution is the region below the line y = x/2 - 1, above y = -4 and towards left of x = 2. (All bold)

4. The solution is the region above the lines y = -2x/3 -7/3 and y = x/3 - 28/3 and below the line y = -x/6 - 13/3. (All bold)

Step-by-step explanation:

3) 4y ≥ -16

y ≥ -4

2x ≤ 4

x ≤ 2

x-2y ≥ 6

2y ≤ x - 2

y ≤ x/2 - 1

The solution is the region below the line y = x/2 - 1, above y = -4 and towards left of x = 2. (All bold)

4) 2x+3y ≥ -7

3y ≥ -2x -7

y ≥ -2x/3 -7/3

x-3y ≤ 28

3y ≥ x - 28

y ≥ x/3 - 28/3

x+6y ≤ -26

6y ≤ -x - 26

y ≤ -x/6 - 13/3

The solution is the region above the lines y = -2x/3 -7/3 and y = x/3 - 28/3 and below the line y = -x/6 - 13/3. (All bold)

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

=====================================

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).

if she was to only spend a dollar she can get 1lb

step-by-step explanation: