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3 years ago

15

If light travels about 3 x 10^5 kilometers per second, about how many kilometers does it travel in one day? Write the answer in

scientific notation.

1 answer:

Dovator [93]3 years ago

6 0

<h2>Answer: 2.59 × 10¹⁰ kilometers/day</h2>

<h3>Step-by-step explanation:</h3>

Since there are 86400 seconds / day

& light travels at 3 x 10⁵ kilometers / second

Then kilometers the light travels in a day

= (86400 seconds / day) × (light travels at 3 x 10⁵ kilometers / second)

= 2.59 × 10¹⁰ kilometers/day

<u></u>

<u></u>

<u>Note:</u>

<em>We know that we have to multiply because the unit 'seconds per day' has the seconds in the numerator, while the unit 'kilometer per second' has the seconds in the denominator, so to cancel we would have to multiply. This gives us the unit we are looking for 'kilometeres per day'.</em>

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Answer:

a) the common difference is 20

b) $x_8=115 , x_{12}=195$

c) the common difference is -13

d) $a_{12}=52, a_{15}=13$

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

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We know common difference d= 20, we need to find $x_1$

Using $x_3=16$ we can find $x_1$

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Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

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The formula used is: $a_n=a_1+(n-1)d$

We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

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$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

So, $\mathbf{a_{12}=52}$

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$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

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