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wariber [46]
3 years ago
12

Does anyone know how to do this ?

Mathematics
2 answers:
lubasha [3.4K]3 years ago
8 0

Answer:

see below

Step-by-step explanation:

Find the circumference

C = 2 * pi *r

C = 2(.23) * pi

C = .46pi

Since the wheel rotated 829 times, multiply the circumference by 829

829 * .46 pi

381.34 pi

If  pi = 3.14

1197.4076 meters

If using the pi button

1198.01494 meters

Alexeev081 [22]3 years ago
3 0

Step-by-step explanation:

The circumference of a wheel is 0.46π, and we simply have to multiply that by 829 which is 381.34π or 1197.41.

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Prove the function f: R- {1} to R- {1} defined by f(x) = ((x+1)/(x-1))^3 is bijective.
Eduardwww [97]

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See explaination

Step-by-step explanation:

given f:R-\left \{ 1 \right \}\rightarrow R-\left \{ 1 \right \} defined by f(x)=\left ( \frac{x+1}{x-1} \right )^{3}

let f(x)=f(y)

\left ( \frac{x+1}{x-1} \right )^{3}=\left ( \frac{y+1}{y-1} \right )^{3}

taking cube roots on both sides , we get

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

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\Rightarrow xy-x+y-1=xy+x-y-1

\Rightarrow -x+y=x-y

\Rightarrow x+x=y+y

\Rightarrow 2x=2y

\Rightarrow x=y

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let y\in R, such that f(x)=\left ( \frac{x+1}{x-1} \right )^{3}=y

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\Rightarrow x+1=\sqrt[3]{y} x- \sqrt[3]{y}

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\Rightarrow x\left ( \sqrt[3]{y} -1 \right ) =1+ \sqrt[3]{y}

\Rightarrow x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

for every y\in R-\left \{ 1 \right \}\exists x\in R-\left \{ 1 \right \} such that x=\frac{\sqrt[3]{y}+1}{\sqrt[3]{y}-1}

Hence f is onto

since f is both one -one and onto so it is a bijective

8 0
3 years ago
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