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Marat540 [252]
3 years ago
15

Lin’s road trip is 75 miles long. She has completed 20% of a trip so far. How many miles has Lin gone so far?

Mathematics
1 answer:
Lynna [10]3 years ago
7 0

You're basically told that Lin has traveled 20% of 75. Since 20% is one fifth, she has traveled 75/5=15 miles so far.

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For the pet the fraction is 2/3 because there is a fish,cat and a dog only cats and dogs have 4 legs and the fraction for the toy or the book is 1/2 because there are 2 objects and she can only get 1
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Can I please get some help on this...I have tried to many times...
Bezzdna [24]

Answer: A

Step-by-step explanation:

First, the problem is g(f(x)). You would plug in f(x) wherever you see an x in g(x). To find the domain, you take the bottom function, and set it equal to 0.

\sqrt{x-2} =0

When you solve that, you get x=2. You know your domain is x≥2, but there is as asymptote at x=11. That means the graph never reaches x=11, but gets very close. You find that by setting the entire equation equal to 0 and solve from there.

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Reika [66]

9514 1404 393

Answer:

  • 0 ≤ m ≤ 7
  • 0.4541 cm/month; average rate of growth over last 4 months of study

Step-by-step explanation:

<u>Part A</u>:

The study was concluded after 7 months. The fish cannot be expected to maintain exponential growth for any significant period beyond the observation period. A reasonable domain is ...

  0 ≤ m ≤ 7

__

<u>Part B</u>:

The y-intercept is the value when m=0. It is the length of the fish at the start of the study.

__

<u>Part C</u>:

The average rate of change on the interval [3, 7] is given by ...

  (f(7) -f(3))/(7 -3) = (4(1.08^7) -4(1.08^3))/4 = 1.08^3·(1.08^4 -1)

   ≈ 0.4541 cm/month

This is the average growth rate of the fish in cm per month over the period from 3 months to 7 months.

3 0
2 years ago
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.

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