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pav-90 [236]
3 years ago
12

Find n so that number sentence below is true and explain your answer, referencing the relevant properties of exponents

Mathematics
1 answer:
zloy xaker [14]3 years ago
3 0

Ugh, I hate FLVS. I'm not exactly sure, but i know that 2 to the 12 power is 24, and 2 to the fourth power is 8. Sorry I'm not being very helpful.

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What is 0.624 an expanded form
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Answer: 0 + 0.6  + 0.02  + 0.004

A Number in expanded form is written like a sentence.

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What is the inverse of the function f(x) = 2x + 1?<br><br> h(x) = one-halfx – one-half
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L. amamammamamammanama

3 1
3 years ago
<img src="https://tex.z-dn.net/?f=%20%20%5Crm%5Csum%20%5Climits_%7Bn%20%3D%200%7D%5E%7B%20%5Cinfty%20%7D%20%20%5Carcsin%20%5Clar
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}

3 0
2 years ago
Whats the distance between point A (-2,-1) and point B (3,2)
Wewaii [24]

Answer:

d = √34

Step-by-step explanation:

(-2, -1) & (3, 2)

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

d = √[(3 + 2)² + (2 + 1)²]

d = √[(5)² + (3)²]

d = √(25 + 9)

d = √34

4 0
3 years ago
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