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zmey [24]
3 years ago
5

The speed of light is 186000 miles per second about how many miles does light travel in an hour

Mathematics
1 answer:
den301095 [7]3 years ago
5 0
One hour is equivalent to 60 minutes and 1 minute is equivalent to 60 seconds, therefore:

60 minutes/hr * 60 seconds/minute = 3600 seconds/hr

At a rate of 186000 miles per second, light will travel 186000*3600 = 669,600,000 miles, which can be restated as 6.7*10^8 miles
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Had 24 beads he divided them into four equal groups and gave one group to his sister how many beads did Lawrence have then
const2013 [10]

Answer:

If Lawrence has 24 beads and divided them into 4 groups you want to see how many he put into each group first

To do this we are going to take 24/4=6

Each group has 6 in it

If he gave one group to his sister, than he gave 6 beads to his sister

Now we want to take 6 away from 24 to find out how many he has left

24-6=18

He now has 18 beads

Hope this helps ;)

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3 years ago
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Use the table to answer the question. City Annual Rainfall San Francisco, CA 20 " Seattle, WA 37 " Miami, FL 55 " New York, NY 4
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2 years ago
Write an equation of the line that has a slope of 2 and contains the point (1, 1) in point-slope form.
exis [7]
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y-y₀=m(x-x₀)

Data:
(1,1)
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7 0
3 years ago
The rectangular walkway will be 3 feet wide and 18 feet long. each 2-foot by 3-foot stone covers an area of 6 square feet. how m
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7 0
3 years ago
81 POINTS
Jobisdone [24]

Base Case: plug in n = 1 (the smallest positive integer)

If n = 1, then 3n-2 = 3*1-2 = 1. Square this and we see that (3n-2)^2 = 1^2 = 1

On the right hand side, plugging in n = 1 leads to...

n*(6n^2-3n-1)/2 = 1*(6*1^2-3*1-1)/2 = 1

Both sides are 1. So that confirms the base case.

-------------------------------

Inductive Step: Assume that

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

is a true statement for some positive integer k. If we can show the statement leads to the (k+1)th case being true as well, then we will have sufficiently proven the overall statement to be true by induction.

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 = k*(6k^2-3k-1)/2

1^2 + 4^2 + 7^2 + ... + (3k-2)^2 + (3(k+1)-2)^2 = (k+1)*(6(k+1)^2-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3(k+1)-2)^2 = (k+1)*(6(k^2+2k+1)-3(k+1)-1)/2

k*(6k^2-3k-1)/2 + (3k+3-2)^2 = (k+1)*(6k^2+12k+6-3k-3-1)/2

k*(6k^2-3k-1)/2 + (3k+1)^2 = (k+1)*(6k^2+9k+2)/2

k*(6k^2-3k-1)/2 + 9k^2+6k+1 = (k+1)*(6k^2+9k+2)/2

(6k^3-3k^2-k)/2 + 2(9k^2+6k+1)/2 = (k*(6k^2+9k+2)+1(6k^2+9k+2))/2

(6k^3-3k^2-k + 2(9k^2+6k+1))/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3-3k^2-k + 18k^2+12k+2)/2 = (6k^3+9k^2+2k+6k^2+9k+2)/2

(6k^3+15k^2+11k+2)/2 = (6k^3+15k^2+11k+2)/2

Both sides simplify to the same expression, so that proves the (k+1)th case immediately follows from the kth case

That wraps up the inductive step. The full induction proof is done at this point.

7 0
3 years ago
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