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timama [110]
3 years ago
9

The hexagon GIKMPR is regular. The dashed line segments form 30 degree angles.

Mathematics
1 answer:
Gwar [14]3 years ago
8 0

Answer:

The correct option is (D).

Step-by-step explanation:

It is given that GIKMPR is regular hexagon. It means it has 6 vertices.

Since the central angle is 360 degree. Therefore the central angle between two consecutive vertices is

\frac{360^{\circ} }{6}=60^{\circ}

It is given that the dashed line segments form 30 degree angles.

We have rotated the hexagon about O to map PQ to RF. Since P and R are consecutive vertices, therefore the angle between them is 60 degree.

The vertex R is immediate next to the vertex P in clockwise direction.

So if we rotate the hexagon at 60 degree clockwise about O, then we can maps PQ to RF.

360^{\circ}-60^{\circ}=300^{\circ}

Therefore we can also rotate the hexagon at 300 degree counterclockwise about O, then we can maps PQ to RF.

Therefore option D is correct.


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

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}

\Rightarrow (x+1)(y-1)=(x-1)(y+1)

\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

Hence f is one - one

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

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

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

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

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

\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

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