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beks73 [17]
3 years ago
9

On which interval is the average rate of change for bicycle production the greatest? A) 1950 to 1960 B) 1960 to 1970 C) 1970 to

1980 D) 1980 to 2000

Mathematics
1 answer:
Alborosie3 years ago
7 0

Answer:

D) 1980 to 2000

Step-by-step explanation:

Finding the average rate of change in each interval to determine the greatest one.

Production per interval

1950-1960 = (21-11) \ million = 10 million

1960-1970 = (41-21)\ million = 20 million

1970-1980 = (71-41)\ million = 30 million

1980-2000 = (161-71)\ million= 90 million

Rate of change (1950-1960)= \frac{10}{11} \times 100= 90.90\%

Rate of change (1960-1970) = \frac{20}{21} \times 100= 95.23\%

Rate of change (1970-1980)=  \frac{30}{41} \times 100= 73.17\%

Rate of change (1980-2000)= \frac{90}{71} \times 100= 126.76\%

∴ rate of change between 1980 to 2000 is 126.78%

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IrinaVladis [17]

The graph which represents the polynomial function g(x) is:

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Step-by-step explanation:

g(x) = x³ + x² - 17x + 15 is a cubic function because the greatest power

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All graphs represent cubic functions

So to find which graph which represents g(x) we will do these steps

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Now let us find the graph which represents g(x)

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∵ x = 0

∴ g(0) = (0)³ + (0)² - 17(0) + 15

∴ g(0) = 15

∵ g(0) is the y-intercept

∴ The graph intersect the y-axis at point (0 , 15)

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The attached figure is the graph of g(x)

The graph which represents the polynomial function g(x) is:

the 2nd graph in the second line

Learn more:

you can learn more about functions in brainly.com/question/10541435

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The area of the top surface of the wedge is the same as the area of a sector.

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Given that: central angle = 45^{o}, radius = 10 centimetre.

Area of the top surface = Area of a sector = (θ/360^{o})\pir^{2}

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