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vekshin1
3 years ago
12

Are the graphs of the lines in the pair parallel? Explain. y = 3/7x+ 11 –3x + 7y = 13

Mathematics
1 answer:
ankoles [38]3 years ago
7 0
Remember that parallel lines have the same slope.

y=3/7x+11 (in slope intercept form already)

-3x+7y=13 (standard form)

Add 3x to both sides, giving you 7y=3x+13, divide by 7 on both sides, giving 


y=3/7x+13/7

Since both equations' slopes' are the same, their graphs will be parallel.


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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:

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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}

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Arrange the following numbers in increasing order: \begin{align*} A &amp;= \frac{2^{1/2}}{4^{1/6}}\\ B &amp;= \sqrt[12]{128}\vph
prisoha [69]

Answer:

The order is "D, B, A, E, C".

Step-by-step explanation:

The numbers are as follows:

A=\frac{2^{1/2}}{4^{1/6}}\\\\B = \sqrt[12]{128}\\\\C=(\frac{1}{8^{1/5}})^{2}\\\\D = \sqrt{\frac{4^{-1}}{2^{-1}\cdot 8^{-1}}}\\\\E = \sqrt[3]{2^{1/2}}\cdot 4^{-1/4}

Simplify the value of A, B, C, D and E as follows:

A=\frac{2^{1/2}}{4^{1/6}}=\frac{2^{1/2}}{2^{2/6}}=\frac{2^{1/2}}{2^{1/3}}=2^{1/2-1/3}=2^{1/6}=1.1225\\\\B = \sqrt[12]{128}=(128)^{1/12}=(2^{7})^{1/12}=2^{7/12}=1.4983\\\\C=(\frac{1}{8^{1/5}})^{2}=(\frac{1}{(2^{3})^{1/5}})^{2}=(\frac{1}{2^{3/5}})^{2}=\frac{1}{2^{3/5\times2}}=\frac{1}{2^{6/5}}=2^{-6/5}=0.4353\\\\D = \sqrt{\frac{4^{-1}}{2^{-1}\cdot 8^{-1}}}=\sqrt{\frac{2^{-2}}{2^{-1}\cdot 2^{-3}}}=\sqrt{2^{-2+1+3}}=2\\\\E = \sqrt[3]{2^{1/2}}\cdot 4^{-1/4}= (2^{1/2})^{1/3}\cdot 2^{-2/4}=2^{-1/3}=0.7937

Arrange the following numbers in increasing order as follows:

D > B > A > E > C

Thus, the order is "D, B, A, E, C".

7 0
3 years ago
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