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Anuta_ua [19.1K]
3 years ago
8

Inverse function of f(x)=my+b

Mathematics
1 answer:
stealth61 [152]3 years ago
4 0

Answer:

f'(x) =  \frac{y - b}{m}

Step-by-step explanation:

We are to find the inverse of f(x) = my + b

First, put it in the form x = my + b

Secondly, replace y with x to get;

y = mx + b

Thirdly, solve for x to get;

x = \frac{y - b}{m}

Finally, replace your x with f'(x) to get;

f'(x) =  \frac{y - b}{m}

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Bogdan [553]
The answer is no more water.

3/4=9/12

3/4
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9/12=9/12
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alekssr [168]

Answer:

Let the vectors be

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b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

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c = 7i + 2j - k

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( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

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