Which statements about this system of equations are true? check all that apply. - x + 6y = 16 8x - 6y = -2 the x-variable will be eliminated when adding the system of equations. the y-variable will be eliminated when adding the system of equations. the sum of the system of equations is - x + 6ythere is only one solution to the system of equations.

The true statements are:

The y-variable will be eliminated when adding the system of equations

There is only one solution to the system of equations is

Step-by-step explanation:

* Lets explain how to solve the problem

- We use the elimination method to solve the system of the

  linear equation

- The solution is one of three cases

# Exactly one solution ⇒ the 2 lines which represented the equations

  intersect each other at one point

# No solution ⇒ the 2 lines which represented the equations are

  parallel to each other

# Infinite solutions ⇒ the two lines are coincide

- In the system of the linear equations of the problem we have two

 linear equations  -x + 6y = 16 and 8x - 6y = -2

- To solve we must to eliminate one of the two variables

∵ The y's in the two equations have the same coefficients and

   different signs

∴ We add the equations to eliminate y

∴ (-x + 8x) + (6y - 6y) = 16 + -2

∴ 7x = 14 ⇒ divide both sides by 7

∴ x = 2

- Substitute the x in any one of the two equations by 2

∴ -2 + 6y = 16 ⇒ add 2 to both sides

∴ 6y = 18 ⇒ divide both sides by 6

∴ y = 3

∴ The solution of the system of the equations is (2 , 3) ⇒ only one

   solution

- Lets check the statements to find the true statements

# The x-variable will be eliminated when adding the system of

   equations is not true

# The y-variable will be eliminated when adding the system of

   equations is true

# The sum of the system of equations is - x + 6y is not true

# There is only one solution to the system of equations is true

no one knows lol

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).