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Llana [10]
3 years ago
6

2. Find the value of x in the triangle. Round your answer to the nearest tenth of a degree.

Mathematics
2 answers:
Nady [450]3 years ago
3 0
The value of x in the triangle is 61.9°.
eimsori [14]3 years ago
3 0
I think the answer is 61.9°
Good Luck!!!
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Step-by-step explanation:

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Use the drop-down menus to complete the statements. in order for f–1(x) to be a function, the domain of f(x) can be restricted t
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Using concepts of the inverse function, it is found that:

The domain of f(x) has to be restricted to x >= 1. Under this restriction, the domain of f-1(x), would be (2,∞), having the same minimum value as the original function f(x).

<h3>How to find the inverse of an equation or a function?</h3>

To find the inverse, we have to exchange x and y and then isolate y. This can only happen if the function f(x) is injective, that is, each value of the output is related to only one value of the input. Then:

  • The domain of f(x) is the range of the inverse function f-1(x).
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Researching this problem on the internet, we are given an absolute value function with vertex at x = 1, meaning that for the inverse, the domain of f(x) has to be restricted to x >= 1. The domain of the inverse would be the range of f(x), which is (2,∞).

Then:

The domain of f(x) has to be restricted to x >= 1. Under this restriction, the domain of f-1(x), would be (2,∞), having the same minimum value as the original function f(x).

More can be learned about inverse functions at brainly.com/question/8824268

#SPJ1

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\bf ~~~~~~~~~~~~\textit{function transformations}&#10;\\\\\\&#10;% function transformations for trigonometric functions&#10;% templates&#10;f(x)=Asin(Bx+C)+D&#10;\\\\&#10;f(x)=Acos(Bx+C)+D\\\\&#10;f(x)=Atan(Bx+C)+D&#10;\\\\&#10;-------------------

\bf \bullet \textit{ stretches or shrinks}\\&#10;~~~~~~\textit{horizontally by amplitude } A\cdot B\\\\&#10;\bullet \textit{ flips it upside-down if }A\textit{ is negative}\\&#10;~~~~~~\textit{reflection over the x-axis}&#10;\\\\&#10;\bullet \textit{ flips it sideways if }B\textit{ is negative}

\bf ~~~~~~\textit{reflection over the y-axis}&#10;\\\\&#10;\bullet \textit{ horizontal shift by }\frac{C}{B}\\&#10;~~~~~~if\ \frac{C}{B}\textit{ is negative, to the right}\\\\&#10;~~~~~~if\ \frac{C}{B}\textit{ is positive, to the left}\\\\&#10;\bullet \textit{vertical shift by }D\\&#10;~~~~~~if\ D\textit{ is negative, downwards}\\\\&#10;~~~~~~if\ D\textit{ is positive, upwards}

\bf \bullet \textit{function period or frequency}\\&#10;~~~~~~\frac{2\pi }{B}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\&#10;~~~~~~\frac{\pi }{B}\ for\ tan(\theta),\ cot(\theta)

with that template in mind, let's see this one.

\bf f(x)=\stackrel{A}{-4}tan(\stackrel{B}{1}x\stackrel{C}{-\pi })+\stackrel{D}{0}&#10;\\\\\\&#10;\stackrel{period}{\cfrac{2\pi }{B}\implies \cfrac{2\pi }{1}\implies 2\pi }&#10;\\\\\\&#10;\stackrel{phase/horizontal~shift}{\cfrac{C}{B}\implies \cfrac{-\pi }{1}}\implies -\pi \qquad \textit{shifted to the right by }\pi \textit{ units}
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3 years ago
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