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Serjik [45]
3 years ago
10

What is the solution to the equation 5/3b^3-2b^2-5 = 2/b^3-2?

Mathematics
1 answer:
sweet [91]3 years ago
7 0

Answer:

The solution to the equation is:

Option c. b=0 and b=4

Step-by-step explanation:

5 / (3b^3-2b^2-5) = 2 / (b^3-2)

Cross multiplication:

5(b^3-2)=2(3b^3-2b^2-5)

Applying distributive property both sides of the equation to eliminate the parentheses:

5(b^3)-5(2)=2(3b^3)-2(2b^2)-2(5)

Multiplying:

5b^3-10=6b^3-4b^2-10

Passing all the terms to the right side of the equation: Subtracting 5b^3 and adding 6 both sides of the equation:

5b^3-10-5b^3+10=6b^3-4b^2-10-5b^3+10

Adding like terms:

0=b^3-4b^2

b^3-4b^2=0

Getting common factor b^2 on the left side of the equation:

b^2 (b^3/b^2-4b^2/b^2)=0

b^2 (b-4) = 0

Two solutions:

(1) b^2=0

Solving for b: Square root both sides of the equation:

sqrt(b^2)=sqrt(0)

Square root:

b=0

(2) b-4=0

Solving for b: Adding 4 both sides of the equation:

b-4+4=0+4

Adding like terms:

b=4

The solution of the equation is: b=0 and b=4 (Option c)

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A jet travels from chicago to kansas city at a speed of 440 knots. If it takes a plane 3 hr longer to fly from kansas city to ch
steposvetlana [31]

d = distance between the two cities

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t₁ = time taken to travel distance going from chicago to kansas city

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t₁ = d/v₁

t₁ = d/440                                        eq-1


v₂ = average speed while going from kansas city to chicago = 110 knots

t₂ = time taken to travel distance going from kansas city to chicago

time taken to travel distance going from kansas city to chicago is given as

t₂ = d/v₂

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