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3 years ago

What is the average rate of change of the amount of the element over the 10 minute experiment?

Mathematics

1 answer:

Svetllana [295]3 years ago

3 0

Answer:

Average rate of change in amount of the element over 10 minutes is 0.15 per minute.

Step-by-step explanation:

Given question is incomplete; find the attachment of this question attached with the answer.

In this experiment change in amount of the element has been plotted on y-axis and duration (time) on x-axis.

Change in amount of the element during 0 to 10 minutes = 1.5 units

Now the duration of this change in amount = 10 minutes

Therefore, rate of change in amount of the element = $\frac{1.5}{10}$

= 0.15 units per minute

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Answer:

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2 years ago

The surface area of a cube is found using the formula sa=6s^2, where s is theside length if the surface area of a cube is 294 sq

aleksley [76]

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3 years ago

Read 2 more answers

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Answer:

x = 1; y = 6

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x = 8; y = 41

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Step-by-step explanation:

Just substitute the x values

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1 year ago

Write a word problem for the inequality -3.5 < -2.5

nlexa [21]

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3 years ago

1) The population of a town was7652in 2016. The population grows at a rate of 1.6%annually.(a)Use the exponential growth model t

marshall27 [118]

Answer:

1. a) population of the town in 2024 = 8697

2.) the earthquake with a magnitude of 8 is a <u>100 times stronger</u> than an earthquake with a magnitude of 6.

Step-by-step explanation:

1) the population of a town was, $N_{0} = 7652$ in 2016.

a) Exponential growth model is given by $N(t) = N_{0} e^{kt}$ where t is the time in years after 2016.

Year 2016 is when t = 0.

k = growth rate constant = 1.6% = 0.016

i) therefore the population of the town in 2024

= $N = N_{0} e^{kt} = 7652e^{0.016\times8} = 8697$ ( to the nearest whole number)

2.) log (I $\times$ M $\times$ S) = magnitude of earthquake where I is the Intensity of the earthquake and S is the intensity of a standard earthquake

therefore I $\times$ M $\times$ S = $10^{magnitude\hspace{0.1cm}of\hspace{0.1cm}earthquake}$

therefore $\frac{strength\hspace{0.1cm}of\hspace{0.1cm}earthquake\hspace{0.1cm}1 }{strength\hspace{0.1cm}of\hspace{0.1cm}earthquake\hspace{0.1cm}2} = \frac{10^{8} }{10^{6} } = 100$

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3 years ago