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2 years ago

Help me plz!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

erica [24]2 years ago

3 0

Answer:

$y = {x}^{2} - 2$

Step-by-step explanation:

Find a Rule means write a function for the graph. The graph is shaped like a U so the graph will be a quadratic, so the graph must have a term

${x}^{2}$

in it. We are given the vertex of the function. We can use vertex formula

$y = (x - h) {}^{2} + k$

where h is the x value of the vertex and k is the y value of the vertex.

$y = (x - 0) {}^{2} - 2$

$y = {x}^{2} - 2$

So the rule is

${x}^{2} - 2$

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4sin²<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B2%7D" id="TexFormula1" title="\frac{x}{2}" alt="\frac{x}{2}" align="absm

raketka [301]

Answer:

$\displaystyle x=\left \{\frac{2\pi}{3}+2\pi k,\frac{4\pi}{3}+2\pi k, \frac{8\pi}{3}+2\pi k, \frac{10\pi}{3}+2\pi k\right \}k\in \mathbb{Z}$

Step-by-step explanation:

Hi there!

We want to solve for $x$ in:

$4\sin^2(\frac{x}{2})=3$

Since $x$ is in the argument of $\sin^2$ , let's first isolate $\sin^2$ by dividing both sides by 4:

$\displaystyle \sin^2\left(\frac{x}{2}\right)=\frac{3}{4}$

Next, recall that $\sin^2x$ is just shorthand notation for $(\sin x)^2$ . Therefore, take the square root of both sides:

$\displaystyle \sqrt{\sin^2\left(\frac{x}{2}\right)}=\sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}}$

Simplify using $\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ :

$\displaystyle \sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \frac{\sqrt{3}}{\sqrt{4}}=\pm \frac{\sqrt{3}}{2}$

Let $\phi = \frac{x}{2}$ .

<h3><u>Case 1 (positive root):</u></h3>

$\displaystyle \sin(\phi)=\frac{\sqrt{3}}{2},\\\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}$

Therefore, we have:

$\displaystyle \frac{x}{2}=\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\frac{x}{2}=\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z},\\\\\begin{cases}x=\boxed{\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{4\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}$

<h3><u>Case 2 (negative root):</u></h3>

$\displaystyle \sin(\phi)=-\frac{\sqrt{3}}{2},\\\phi = \frac{4\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{5\pi}{3}+2\pi k, k\in \mathbb{Z},\\\begin{cases}x=\boxed{\frac{8\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{10\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}$

8 0

2 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it: