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2 years ago

Solve for x. 4x-4<8 AND 9x+5>23

Mathematics

2 answers:

pogonyaev2 years ago

6 0

4x-4<8

Add 4 on both sides

4x<12

Divide 4 on both sides

x<3

9x+5>23

Subtract 5 on both sides

9x>18

Divide 9 on both sides

x>2

*for reference, I will go ahead and add in that x will be greater than 2 and less than 3

2<x<3

Hope I helped!!

Anna007 [38]2 years ago

4 0

Answer:

2 < x < 3

Step-by-step explanation:

4x - 4 < 8
9x + 5 > 23

Solve for x in the first equation. Add 4 to both sides.

4x < 12

Divide both sides by 4.

x < 3

Solve for x in the second equation. Subtract 5 from both sides.

9x > 18

Divide both sides by 8.

x > 2

We know that x is greater than 2 but less than 3, therefore:

2 < x < 3

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3 years ago

Read 2 more answers

Circle 1, Circle 2, and Circle 3 have the same center and have radii, respectively, of r₁ cm, r₂ cm, r₃ cm, where r₁ < r₂ &lt

ira [324]

Answer:

a) A₁ /A₂ = r₁² / (r₂² - r₁²)

b) A₂ /A₃ = (r₂² - r₁²) / (r₃² - r₂²)

Step-by-step explanation:

We have Circle 1 and area A₁

Area of circle 2 outside circle 1 = A₂

Area of circle 3 outside circle 2 = A₃

On the other hand we have

A₁ = π*r₁² area of circle 1

A₂´ = π*r₂² area of circle 2

A₃´ = π*r₃² area of circle 3

All areas in cm²

a) A₁ /A₂

A₁ = π*r₁²

According to problem statement A₂ = π*r₂² - A₁

A₂ = π*r₂² - π*r₁² ⇒ A₂ = π* (r₂² - r₁²)

Then A₁ /A₂ = π*r₁² / [π* (r₂² - r₁²)]

A₁ /A₂ = r₁² / (r₂² - r₁²)

b) A₂ /A₃

A₂ = π* (r₂² - r₁²)

And

A₃ = π* (r₃² - r₂²)

Therefore

A₂ /A₃ = π* (r₂² - r₁²) / π* (r₃² - r₂²) ⇒ A₂ /A₃ = (r₂² - r₁²) / (r₃² - r₂²)

A₂ /A₃ = (r₂² - r₁²) / (r₃² - r₂²)

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