$\bf cos^{-1}\left( \cfrac{1}{8} \right)=\theta \qquad \qquad cos(\theta )=\cfrac{\stackrel{adjacent}{1}}{\stackrel{hypotenuse}{8}}\impliedby \textit{let's find the \underline{opposite}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{8^2-1^1}=b\implies \pm\sqrt{63}=b ~\hfill tan(\theta )=\cfrac{\stackrel{opposite}{\pm\sqrt{63}}}{\stackrel{adjacent}{1}} \\\\\\ ~\hspace{34em}$

$\bf \stackrel{\textit{keeping in mind that}}{tan\left(cos^{-1}\left( \frac{1}{8} \right) \right)}\implies tan(\theta )$

5 0

2 years ago

Which expression is equivalent to (9x - 3 1/8)-(7x + 1 3/8)

strojnjashka [21]

2x - 4 1/2 is equivalent

7 0

3 years ago

What is the extraneous solution?

8_murik_8 [283]

<h3>Answer:</h3>

x = 1

<h3>Step-by-step explanation:</h3>

<em>The only solution is an extraneous solution</em>, which is to say the equation has no solution.

The rational expression reduces to -1 (for x≠1), which makes the equation ...

1 = 1/x

The only solution to this is x=1, which is specifically disallowed.

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If you subtract the right side, the equation becomes ...

((-x+1)(x) +2(x)(x -1) -(x -1))/(x -1) = 0

(-x^2 +x +2x^2 -2x -x +1)/(x -1) = 0 . . . . eliminate parentheses in the numerator

(x^2 -2x +1)/(x -1) = 0 . . . . . . . . . . . . . . . collect terms

(x -1)^2/(x -1) = 0 . . . . . . . . . . . . . . . . . . . factor

This is undefined for the only value of x that could possibly be a solution, x=1.

8 0

3 years ago

What is the slope for y−3=−51​(x+17)

What is the slope for y−3=−51(x+17)