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AleksAgata [21]
3 years ago
15

Calculate the product in square (m2)

Mathematics
1 answer:
Jet001 [13]3 years ago
4 0

1m=100cm\to1cm=\dfrac{1}{100}m\\\\1dam=10m\\\\\text{therefore}\\\\\dfrac{2}{5}cm=\dfrac{2}{5}\cdot\dfrac{1}{100}m=\dfrac{1}{250}m\\\\\dfrac{1}{3}dam=\dfrac{1}{3}\cdot10m=\dfrac{10}{3}m\\\\\dfrac{2}{5}cm\times\dfrac{1}{3}dam=\dfrac{1}{250}m\times\dfrac{10}{3}m=\dfrac{1}{75}m^2

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Which pair of funtions is not a pair of inverse functions? please help!!
antiseptic1488 [7]

Answer:

f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}

Step-by-step explanation:

we know that

To find the inverse of a function, exchange variables x for y and y for x. Then clear the y-variable to get the inverse function.

we will proceed to verify each case to determine the solution of the problem

<u>case A)</u> f(x)=\frac{x+1}{6} , g(x)=6x-1

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y+1}{6}

Isolate the variable y

6x=y+1

y=6x-1

Let

f^{-1}(x)=y

f^{-1}(x)=6x-1

therefore

f(x) and g(x) are inverse functions

<u>case B)</u> f(x)=\frac{x-4}{19} , g(x)=19x+4

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y-4}{19}

Isolate the variable y

19x=y-4

y=19x+4

Let

f^{-1}(x)=y

f^{-1}(x)=19x+4

therefore

f(x) and g(x) are inverse functions

<u>case C)</u> f(x)=x^{5}, g(x)=\sqrt[5]{x}

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=y^{5}

Isolate the variable y

fifth root both members

y=\sqrt[5]{x}

Let

f^{-1}(x)=y

f^{-1}(x)=\sqrt[5]{x}

therefore

f(x) and g(x) are inverse functions

<u>case D)</u> f(x)=\frac{x}{x+20} , g(x)=\frac{20x}{x-1}

Find the inverse of f(x)

Let

y=f(x)

Exchange variables x for y and y for x

x=\frac{y}{y+20}

Isolate the variable y

x(y+20)=y

xy+20x=y

y-xy=20x

y(1-x)=20x

y=20x/(1-x)

Let

f^{-1}(x)=y

f^{-1}(x)=20x/(1-x)

\frac{20x}{1-x}\neq \frac{20x}{x-1}

therefore

f(x) and g(x) is not a pair of inverse functions

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I think it (3) one I’m not sure
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