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CaHeK987 [17]
3 years ago
13

Worth 30 points

Mathematics
2 answers:
Scilla [17]3 years ago
8 0

Hi Nicole,


Question 34:

2 1/4 pounds of mulch <u>for every</u> 1 1/3 pounds of gravel.

Gravel - 172 / 1 1/3 = 129

Mulch - 172 / 2 1/4 = 76.44444...


Question 35:

a. 18 beats in 15 seconds.

b. 25 beats in 10 seconds.

Part A:

Find beats per minute:

1 minute = 60 seconds

60 / 15 = 4

18 x 4 = 72

Answer - 72 beats per minute (Resting)

150 beats per minute (Running)


Part B:

72 times per minute resting

Find 25 times in 10 seconds in minutes,

1 minute = 60 seconds

60 / 10 = 6

25 x 6 = 150

So while running the friend has 150 beats per minute.

Find the number of beats in 3 minutes for both:

Resting:

72 x 3 = 216

216 beats in 3 minutes

Running:

150 x 3 = 450

450 beats in 3 minutes

Subtract:

450 - 216 = 234

Answer - Your friend's heart beat is <em><u>234 beats </u></em>more in 3 minutes than at rest.


Hope This Helps!

Ivenika [448]3 years ago
3 0

The answer to question 34:

Let 'x' = pounds of gravel. The ratio of mulch to gravel is given 2 1/4 | 1 1/3. Which equals 9/4 | 4/3 = 27/16.



So the total gravel and much is equal to x + 27/16x = 172  


Solving for 'x'



43/16 x = 172    


x = 172 (16/43) = 64 pounds of gravel           Mulch = 27/16x = 64 (27/16)= 108 pounds

The answer to question 35:

A.) 1,080 beats per minute at rest and 1,500 beats per minute while running.

B.) 4,500 (while running) - 3,240 (at rest) = 1,260 more beats while running than at rest.

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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 < ∞.

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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.

_____

<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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3 years ago
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