All categories

3 years ago

If mountains rose by 1 mm/yr, how high would they be (in meters) after 1,000 years? 10 million years? 50 million years?

Mathematics

1 answer:

romanna [79]3 years ago

4 0

Answer:

a)after 1,000 years?

1 meter

b) 10 million years?

10,000 meters

c) 50 million years?

50,000 meters

Step-by-step explanation:

We are told in the question that mountains rise 1 mm per year.

a) After 1000 years

1 year = 1 mm

1000 years =

Cross Multiply

1000 × 1 mm

= 1000mm

Converting to meters

1 mm = 0.001 m

1000 mm =

1000 × 0.001m

= 1 meter

b) After 10 million (10,000,000) years

1 year = 1 mm

10000000years =

Cross Multiply

10000000 × 1 mm

= 10,000,000mm

Converting to meters

1 mm = 0.001 m

10,000,000mm =

10,000,000 × 0.001m

= 10,000 meters

c) After 50 million years(50,000,000)

After 50,000,000 years

1 year = 1 mm

50,000,000years =

Cross Multiply

50,000,000 × 1 mm

= 50,000,000mm

Converting to meters

1 mm = 0.001 m

50,000,000mm =

50,000,000 × 0.001m

= 50,000 meters

You might be interested in

Please help!! will mark branliest

Mashcka [7]

Answer:

a. After the first bounce, the ball will be at 85% of 8 ft. After 2 bounces, it'll be at 85% of 85% of 8 feet. After 3 bounces, it'll be at (85% of) (85% of) (85% of 8 feet). You can see where this is going. After n bounces the ball will be at

$h(n) = (85\%)^n 6ft = (\frac{17}{20})^n(6ft\times 12\frac{in}{ft}) = (\frac{17}{20})^n\times72'$

b. After 8 bounces we can apply the previous formula with n = 8 to get

$h(8)=(\frac{17}{20})^8\times 72' = 19.62'$

c. The solution to this point requires using exponential and logarithm equations; a more basic way would be trial and error using the previous $h(n)$ increasing the value of n until we find a good value. I recommend using a spreadsheet for that; the condition will lead to the following inequality:

$1>(\frac{17}{20})^n\times 72$ Let's first isolate the fraction by dividing by 72.

$(\frac{17}{20})^n$ Now, to get numbers we can plug in a calculator, let's take the natural logarithm of both sides:

$ln ((\frac{17}{20})^n) < ln \frac 1{72} \rightarrow n\times ln (\frac{17}{20}) < -ln72$ . Now the two quantities are known - or easy to get with any calculator, replacing them and solving for n we get:

$-(0.16)n < - 4.27 \rightarrow n>26.31$ Now, since n is an integer - you can't have a fraction of a bounce after all, you pick the integer right after that, or n>27.