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Leona [35]
3 years ago
12

What is the inverse function of f(x)= x/x-2

Mathematics
1 answer:
defon3 years ago
7 0
Answer: f^-1(x) = 2x/x-1

Explanation:
Setting y=f(x)

y=x/x-2

This may be rearranged as follows

y(x-2) = x

x•y-2y = x

x•y-x = 2y

x=2y/y-1

The expression shows x in terms of y.

That is it! The inverse function of f(x).

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That would be $16.50
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What is the domain of the square root function graphed below?
lesya [120]

Answer:

x \geqslant 0

Option D is the right option.

Explanation:

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4 0
3 years ago
Simplify the following​
den301095 [7]

Answer:

1) 11\sqrt{3}

2) 2\sqrt{2}

3) 20\sqrt{3}  + 15\sqrt{2}

4) 53 + 12\sqrt{10}

5) -2

6) 7\sqrt{2}  - 5\sqrt{3}

Step-by-step explanation:

1) 2\sqrt{12} + 3\sqrt{48} - \sqrt{75}

=(2 × 2\sqrt{3} )+ (3 × 4\sqrt{3}) - 5\sqrt{3}

= 4\sqrt{3} + 12\sqrt{3} - 5\sqrt{3}

= 11\sqrt{3}

2) 4\sqrt{8} -2\sqrt{98} + \sqrt{128}

= (4 × 2\sqrt{2}) - (2 × 7\sqrt{2}) + 8\sqrt{2}

= 8\sqrt{2} - 14\sqrt{2} +8\sqrt{2}

= 2\sqrt{2}

3) 5\sqrt{12\\} - 3\sqrt{18} + 4 \sqrt{72}  +2\sqrt{75}

= 5× 2\sqrt{3} - 3×3\sqrt{2} + 4×6\sqrt{2} + 2×5\sqrt{3}

= 10\sqrt{3} - 9\sqrt{2} +24\sqrt{2} +10\sqrt{3}

= 20\sqrt{3}  + 15\sqrt{2}

4) (2\sqrt{2}  + 3\sqrt{5} )^{2}

= 8 + 12\sqrt{10} + 45

= 53 + 12\sqrt{10}

5) (1+\sqrt{3} ) (1-\sqrt{3} )

= 1 - 3

= -2

6) (2\sqrt{6} -1) (\sqrt{3} -\sqrt{2}  )

= 2\sqrt{18}-2\sqrt{12}  -\sqrt{3}  +\sqrt{2}

= 2×3\sqrt{2} - 2×2\sqrt{3} - \sqrt{3} + \sqrt{2}

= 6\sqrt{2}  - 4\sqrt{3} -\sqrt{3} +\sqrt{2}

= 7\sqrt{2}  - 5\sqrt{3}

Hope the working out is clear and will help you. :)

5 0
2 years ago
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Answer:

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