HELP ASAP PLEASE {will gave thanks and a brainliest}

Mathematics

1 answer:

Luba_88 [7]3 years ago

8 0

Alright here I go:

9.98 * 10^6 in standard form is 9,980,000
7.3 * 10^7 in standard form is 73,000,000
73,000,000 subtracted by 9,980,000 is 63020000.
63020000 in scientific notation is 6.302 * 10^7
Hope I helped!

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sineoko [7]

Recall that over an appropriate domain,

$\arcsin(x) \pm \arcsin(y) = \arcsin\left(x \sqrt{1-y^2} \pm y \sqrt{1-y^2}\right)$

Let $x=\frac1{n+1}$ and $y=\frac1{n+2}$ (these belong to the "appropriate domain", so the identity holds). We have

$\dfrac{\sqrt{n+3}}{(n+2)\sqrt{n+1}} = \dfrac1{n+1} \sqrt{\dfrac{(n+3)(n+1)}{(n+2)^2}} = \dfrac1{n+1} \sqrt{1 - \dfrac1{(n+2)^2}} = x \sqrt{1-y^2}$

and

$\dfrac{\sqrt n}{(n+1)\sqrt{n+2}} = \dfrac1{n+2} \sqrt{\dfrac{n(n+2)}{(n+1)^2}} = \dfrac1{n+2} \sqrt{1 - \dfrac1{(n+1)^2}} = y \sqrt{1-x^2}$

Then the sum telescopes, as

$\displaystyle \sum_{n=0}^\infty \arcsin\left(x \sqrt{1-y^2} - y \sqrt{1-x^2}\right) = \sum_{n=0}^\infty \left( \arcsin(x) - \arcsin(y) \right) \\\\ = \left(\arcsin(1) - \arcsin\left(\frac12\right)\right) + \left(\arcsin\left(\frac12\right) - \arcsin\left(\frac13\right)\right) + \cdots \\\\ = \arcsin(1) = \boxed{\frac\pi2}$