Choose a relationship model that will reach $0 in the same number of days as this scenario: mika has $12 and spends $2 each day. y = –2x – 12 y = –3x + 18 y = –4x + 12 y = –12x + 2

Model (2) has the same number of days as Mika

Step-by-step explanation:

- The relationship models are:

→ y = -2x - 12 ⇒ (1)

→ y = -3x + 18 ⇒ (2)

→ y = -4x + 12 ⇒ (3)

→ y = -12x + 2 ⇒ (4)

- We need to find the relationship model that will reach $0 in the

  same number of days as this scenario:

   Mika has $12 and spends $2 each day

- The rate of spends of Mika is $2 per day

∵ x is the number of days

∵ y is the amount of money after x days

∵ Mika has $12

∵ Mika spends $2 each day

∴ y = -2x + 12

- We need to know the amount of money will be $0 after how

  many days

- Put y = 0

∴ 0 = -2x + 12

- Subtract 12 from both sides

∴ -12 = -2x

- Divide both sides by -2

∴ x = 6

∴ Mika will have $0 after 6 days

- Let us do the same in each model to find x and then find which of

   them have the same x as Mika

→ Model (1)

∵ 0 = -2x - 12

- Add 12 to both sides

∴ 12 = -2x

-Divide both sides by -2

∴ x = -6

→ Model (2)

∵ 0 = -3x + 18

- Subtract 18 from both sides

∴ -18 = -3x

-Divide both sides by -3

∴ x = 6

→ Model (3)

∵ 0 = -4x + 12

- Subtract 12 from both sides

∴ -12 = -4x

-Divide both sides by -4

∴ x = 3

→ Model (4)

∵ 0 = -12x + 2

- Subtract 2 from both sides

∴ -2 = -12x

-Divide both sides by -12

∴ x = 1/6

* Model (2) has the same number of days as Mika

step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)