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navik [9.2K]
3 years ago
8

The hypotenuse of a right

Mathematics
1 answer:
Lina20 [59]3 years ago
8 0
None I would think because none of them are them because if you it is not the
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"y-axis, x-axis, y-axis, x-axis" is the set of reflections among the following choices given in the question that <span>would carry parallelogram ABCD onto itself. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that this is the answer that has helped you.</span>
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ACB

Step-by-step explanation:

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Could you please help me for this question?
Olin [163]

Answer:

  See attached for graphs

  g(x) -- domain: -∞ < x < ∞; range: 0 < y < ∞

  g^-1(x) -- domain: 0 < x < ∞; range: -∞ < y < ∞

Step-by-step explanation:

g(x) is an exponential decay function. Its base is 1/3, so each increase of 1 unit in x will multiply the y-value by a factor of 1/3. The graph will rapidly approach its horizontal asymptote of y=0 as x gets large. The y-intercept is (0, 1). Just as y gets smaller as x increases, so it gets larger as x decreases. Each decrease of x by 1 unit causes the y-value to be multiplied by 3.

__

The graph of g^-1(x) is the graph of g(x) reflected across the line y=x. That is, each coordinate pair (x, y) on the graph of g(x) becomes a point (y, x) on the graph of the inverse function. In order to graph g^-1(x), you don't need to write down the function, you only need to know the relationship between the graphs.

Just as x- and y- are interchanged on the graph, so the domain, range, and intercepts are interchanged. g^-1(x) will have a vertical asymptote of x=0, and an x-intercept of (1, 0). The domain of g^-1(x) is the range of g(x): 0 < x < ∞; and the range of g^-1(x) is the domain of g(x): -∞ < y < ∞.

__

The attached graph shows g(x) in red and g^-1(x) in blue. As you can see, we created the graph simply by interchanging x and y. The line y=x is shown for reference, so you can see that each curve is a reflection of the other across that line.

_____

<em>Additional comment</em>

The explicit expression for g^-1(x) can be found by solving for y:

  x = g(y)

  x=\left(\dfrac{1}{3}\right)^y=\dfrac{1}{3^y}=3^{-y}\\\\ \log(x)=-y\cdot\log(3)\qquad\text{take logarithms}\\\\y=-\dfrac{\log{x}}{\log{3}}=-\log_3{x}\qquad\text{use the change of base relation}\\\\\boxed{g^{-1}(x)=-\log_3{x}}

If you're familiar with the log function, you know it has an x-intercept of 1 and a vertical asymptote at x=0. The base of the log function is simply a vertical scale factor. The minus sign reflects it across the x-axis.

6 0
2 years ago
What is the solution to the equation fraction 1 over 4x = 5?
AVprozaik [17]
\displaystyle  \frac{1}{4x}=5\quad \Big|\cdot 4x\\\\1=20x\quad\Big|:20\\\\x= \frac{1}{20}
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3 years ago
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