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laila [671]
3 years ago
8

The school gym can hold 650 people. A permanent bleacher in the gymn holds 136 people. The event organizers are setting up 25 ro

ws with an equal number of chairs. At most, how many chairs can be in each row?
Mathematics
1 answer:
igomit [66]3 years ago
3 0
650 -136=514
514 divided by 25 = 20.56
So it would be 21
This will give you a little extra chairs but they have to be equal in each row
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Solving Rational Functions Hello I'm posting again because I really need help on this any help is appreciated!!​
Greeley [361]

Answer:

x = √17 and x = -√17

Step-by-step explanation:

We have the equation:

\frac{3}{x + 4}  - \frac{1}{x + 3}  = \frac{x + 9}{(x^2 + 7x + 12)}

To solve this we need to remove the denominators.

Then we can first multiply both sides by (x + 4) to get:

\frac{3*(x + 4)}{x + 4}  - \frac{(x + 4)}{x + 3}  = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}

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Now we can multiply both sides by (x + 3)

3*(x + 3)  - \frac{(x + 4)*(x+3)}{x + 3}  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

3*(x + 3)  - (x + 4)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}

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(2*x + 5)*(x^2 + 7x + 12)  = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)

(2*x + 5)*(x^2 + 7x + 12)  = (x + 9)*(x + 4)*(x+3)

Now we need to solve this:

we will get

2*x^3 + 19*x^2 + 59*x + 60 =  (x^2 + 13*x + 3)*(x + 3)

2*x^3 + 19*x^2 + 59*x + 60 =  x^3 + 16*x^2 + 42*x + 9

Then we get:

2*x^3 + 19*x^2 + 59*x + 60 - (  x^3 + 16*x^2 + 42*x + 9) = 0

x^3 + 3x^2 + 17*x + 51 = 0

So now we only need to solve this.

We can see that the constant is 51.

Then one root will be a factor of 51.

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But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.

This means that we can take a factor (x + 3) out, so we can rewrite our equation as:

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The other two solutions are when the other term is equal to zero.

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And neither of these have problems in the denominators, so we can conclude that the solutions are:

x = √17 and x = -√17

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