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love history [14]

3 years ago

8

I urgently need help!!

1 answer:

STALIN [3.7K]3 years ago

5 0

Answer:

<h3> $4.9 ~cm$ </h3>

❅❅❅❅❅❅❅

<h3>✰<u>Using the intersecting chord theorem:-</u></h3>

$x(10)=7(7)$

$10x=49$

$x=49/10$

$=49$

✩★✩★✩★✩★✩★✩

hope it helps...

have a great day!!

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Create a quadratic equation that can be solved by factoring. Give the factored answer.

elixir [45]

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saul85 [17]

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

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

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

We can also write it as:

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Squaring both sides, we get:

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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>

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Therefore, the value of the expression is 3.

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$\sf\longmapsto x=3$

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$\sf\longmapsto x = \sqrt{9}$

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

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

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

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

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

This pattern will continue.

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

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