SOLUTION: One common use of positive and negative exponents is in writing numbers as scientific notation. for example 0.000052 can be writtrn as 5.2*10^-5. A) Use scientific notation and the

Algebra ->  Decimal-numbers -> SOLUTION: One common use of positive and negative exponents is in writing numbers as scientific notation. for example 0.000052 can be writtrn as 5.2*10^-5. A) Use scientific notation and the      Log On


   



Question 170841: One common use of positive and negative exponents is in writing numbers as scientific notation. for example 0.000052 can be writtrn as 5.2*10^-5. A) Use scientific notation and the law of exponents to evaluate (0.000054)(0.00000033)(0.013). Leave your answer in scientific notation. B) If a space probe travels at 12 kilometers per second for 10 years, how far will it travel? Write your answer in scientific notation. ( Assume 1 year =365)
Scientific Notation: A number is expressed in scientific notation when it is in the form a*10^n, where a is greater than or equal to 1 but less than 10 and n is an integer.
I got the first part, I think.
5.4*10^-5
3.3*10^-7
1.3*10^-2= 23.17*10^-14
The second part I have no clue.

Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A) Use scientific notation and the law of exponents to evaluate
(0.000054)(0.00000033)(0.013)= 5.4x10^-5 * 3.3x10^-7 * 1.3x10^-2
= 2.3166x10^-13
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. Leave your answer in scientific notation.
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B) If a space probe travels at 12 kilometers per second for 10 years, how far will it travel? Write your answer in scientific notation. ( Assume 1 year =365)
Find # of seconds in 10 years:
60*60 = 3.6x10^3 # of seconds in one hour
3.6x10^3*24*365*10*12 = 3.78432x10^9 km/10yrs
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Cheers,
Stan H.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
A)

You are doing good. But to get the answer in scientific notation, the first number in front of the "10 to the exponent" MUST be in between 0 and 10. So the number 23.17*10^-14 must become 2.3*10^-13 since I shifted the decimal 1 spot to the left.

So the answer is 2.3*10^-13





B)

First, we need to convert 10 years into seconds (we could go the other way, but it might be confusing)


Since we're assuming that there are 365 days in a year, multiply 365 by 10 to get

365*10=3650


So there are 3,650 days in 10 years


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Now because there are 24 hours in 1 day, multiply 24 by 3,650 to get


24*3,650 = 87,600

So there are 87,600 hours in 10 years.

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Since there are 60 minutes in an hour and 60 seconds in a minute, there are 60*60=3600 seconds in an hour


Now simply multiply 3,600 by 87,600 to get

3,600*87,600 = 315,360,000



So there are 315,360,000 seconds in a year


Since t=10 years, this means that t=315360000 seconds which in scientific notation is t=3.15%2A10%5E8

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d=rt Start with the distance rate time formula


d=12%283.15%2A10%5E8%29 Plug in r=12 and t=3.15%2A10%5E8


d=37.8%2A10%5E8 Multiply


d=3.78%2A10%5E9 Shift the decimal one spot to the left (this will raise the exponent by 1)


So the solution is d=3.78%2A10%5E9 km which is 3,780,000,000 km. In words, this distance is 3 billion, 780 million kilometers.