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Dominik [7]
3 years ago
14

Simplify the expression given below.

Mathematics
1 answer:
trasher [3.6K]3 years ago
5 0

Answer:

A) 1/(3x+1)

Step-by-step explanation:

(x+2)/(x³+2x²-9x-18)÷(3x+1)/(x²-9)

(x+2)/(x+2)(x+3)(x-3)÷(3x+1)/(x+3)(x-3)

since it’s dividing by a fraction, invert and multiply

(x+2)/(x+2)(x+3)(x-3) times (x+3)(x-3)/(3x+1)

The (x+2)’s cancel out.

The (x+3)’s cancel out.

The (x-3)’s cancel out.

You are left with 1/(3x+1)

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PLEASE HELP 50 POINTS!!
sergij07 [2.7K]

Answer:

This is not even 50 points?

Step-by-step explanation:

3 0
2 years ago
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Nathan shares out 12 sweets he gives yasmin 1 sweet for 3 sweets he buys how many sweets does nathan get
Yuri [45]

Given that, Nathan shares out 12 sweets, if he gives Yasmin 1 sweet for 3 sweets he buys, Nathan will get 9 sweets.

<h3>How many sweets does Nathan gets?</h3>

Given that, Nathan shares out 12 sweets, he gives Yasmin 1 sweet for 3 sweets he buys.

Let the sweet be represented by x

For each x sweet for Yasmin, Nathan gets 3x sweets

Hence

x + 3x = 12

We solve for x

4x = 12

x = 12/4

x = 3

Hence;

Yasmin gets x sweet = 3

Nathan gets 3x sweets = 3 × 3 = 9

Given that, Nathan shares out 12 sweets, if he gives Yasmin 1 sweet for 3 sweets he buys, Nathan will get 9 sweets.

Learn more word problems here: brainly.com/question/14539651

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6 0
2 years ago
Write and equivalent expression to the expression below (17x93) (17x97)
mart [117]
(17 * 93) ^ 2 = 1581
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Which vector below goes from (0,0) to (2,2)? <br><br> A. c<br><br> B. a<br><br> C. b<br><br> D. d
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Explain why a quadratic equation with a positive discriminant has two real solutions, a quadratic equation with a negative discr
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Answer:

A quadratic equation can be written as:

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where a, b and c are real numbers.

The solutions of this equation can be found by the equation:

x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}

Where the determinant is D = b^2 - 4*a*c.

Now, if D>0

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√D = R

then the solutions are:

x = \frac{-b +- R }{2*a}

Where each sign of R is a different solution for the equation.

If D< 0, we have the square root of a negative number, then we have a complex component:

√D = i*R

x = \frac{-b +- C*i }{2*a}

We have two complex solutions.

If D = 0

√0 = 0

then:

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We have only one real solution (or two equal solutions, depending on how you see it)

3 0
3 years ago
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