What is the slope of a line parallel to the line whose equation is 3x−2y=18. Fully reduce your answer.
2 answers:
You might be interested in
Interpreting the situation, it is found that Marjorie will have 240 left after paying her penalty.
---------------------
- When the piece lands on the red square, the penalty is equal to <u>30 times the sum they roll on two numbers.</u>
- Marjorie rolls 4 and 2, thus, the sum is 6.
- The penalty is of
- She currently has 420, the penalty removes 180, thus, she will have 420 - 180 = 240 left after paying her penalty.
A similar problem is given at brainly.com/question/12166019
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)
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.
