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RideAnS [48]
4 years ago
11

For a function p(x)= x^2-9, what is the inverse function for the domain [0, infinity\

Mathematics
1 answer:
kkurt [141]4 years ago
7 0
An inverse function is a function that reverses another function, so we have a function called:

p(x)

Then, the inverse function will be as follows:

p^{-1}(x)

Given that y = p(x), we need to isolate x in terms of y:

y =  x^{2} -9
∴ y+9 = x^{2}

So:

x=+\sqrt{y+9} and x=-\sqrt{y+9}

therefore, exchanging variables x and y:

y=+\sqrt{x+9} and y=-\sqrt{x+9}
p^{-1}(x)=+\sqrt{x+9} and p^{-1}(x)=-\sqrt{x+9}

Which are the figures shown below.

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I need help on this asap please
Olin [163]

Answer:

450m^2

15m x 10m = 150m^2

20m x 30m = 600m^2

600m^2 - 150m^2 = 450m^2

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Ket [755]

ANSWER

34 {units}^{2}

EXPLANATION

The width of the rectangle is |AB|

A(-1,4) and B(3,3)

We use the distance formula,

d =  \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }

This implies that,.

d =  \sqrt{ {(3- - 1)}^{2} + {(3 - 4)}^{2} }

d =  \sqrt{ {(4)}^{2} + {( - 1)}^{2} }

d =  \sqrt{ 16+ 1 }

d = \sqrt{17}

The length is BC

C(1,-5) and B(3,3)

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d =  \sqrt{ {2}^{2} + {(8)}^{2} }

d =  \sqrt{4 + 64 }

d = \sqrt{68}

d = 2 \sqrt{17}

The area is

= l \times w

= 2 \sqrt{17}  \times  \sqrt{17}

2 \times 17 = 34 {units}^{2}

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