An equation is shown below:

1 answer:

8 0

6(2x-11)+15=3x+12 Given
12x-66+15=3x+12 Distribution
12x-51=3x+12 Combine like terms
12x=3x+63 addition
9x=63 subtraction
x=7 division

the value of x that makes the equation true is 7.

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A cash register contains $10 bills and $100 bills with a total value of $2010. If there are 30 bills total, then how many of eac

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Answer:

there will be 19 $100 bills

and 11 $10 bills

Step-by-step explanation:

$100*19=1900\\10*11=110\\1900+110=2010$

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What's the domaing and range of this (sorry for the quality, its the best I could get)

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Answer:

domain: [-3,2]

range: [-3,2]

Step-by-step explanation:

domain: all possible x values *maximum values of x

range: all possible y values *maximum values of x

brackets means the max while parenthesis mean infinite

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Find an equation of the line passing through the points (2, 2) and (-1,-6)?

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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.

_____

Additional comment

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.