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asambeis [7]
4 years ago
14

A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one

black ball is to be included in the draw?
Mathematics
1 answer:
Mandarinka [93]4 years ago
6 0

Answer: 96

Step-by-step explanation:

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\large\underline{\sf{Solution-}}

<u>Let us assume that:</u>

\sf \longmapsto x =  \sqrt{6 +  \sqrt{6 +  \sqrt{6 + ... \infty } } }

We can also write it as:

\sf \longmapsto x =  \sqrt{6 +  x }

Squaring both sides, we get:

\sf \longmapsto  {x}^{2}  =6 +  x

\sf \longmapsto  {x}^{2} - x - 6  =0

By splitting the middle term:

\sf \longmapsto  {x}^{2} - 3x + 2x - 6  =0

\sf \longmapsto x(x - 3) + 2(x - 3 ) =0

\sf \longmapsto (x+ 2)(x - 3 ) =0

<u>Therefore:</u>

\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}

\sf \longmapsto x =  - 2,3

<u>But x cannot be negative. </u>

\sf \longmapsto x = 3

Therefore, the value of the expression is 3.

\large\underline{\sf{Verification-}}

Given:

\sf\longmapsto x=3

We can also write it as:

\sf\longmapsto x = \sqrt{9}

\sf\longmapsto x = \sqrt{6+3}

\sf\longmapsto x = \sqrt{6 + \sqrt{9}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}

This pattern will continue.

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}

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Step-by-step explanation:

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