Rearrange the equation 5x-3y+9=0 from standard form to slope/y-intercept form.
1 answer:
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First you need to find the area of the whole circle.
The radius is given as 10, so just insert 10 into the equation of the area of a circle equation.
Putting
after the 100 because that is what is commonly done.
Because the circle is then being split into 4 equal sections, we need to find 1/4 of the area of the whole circle to find the area of one of these sections.
Since the area of the whole circle is
then we just divide that by 4,
leaving us with 25 and that is our final answer
The radius is given as 10, so just insert 10 into the equation of the area of a circle equation.
Putting
Because the circle is then being split into 4 equal sections, we need to find 1/4 of the area of the whole circle to find the area of one of these sections.
Since the area of the whole circle is
leaving us with 25
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.
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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 < ∞.
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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.
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<em>Additional comment</em>
The explicit expression for g^-1(x) can be found by solving for y:
x = g(y)
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.
