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bonufazy [111]
3 years ago
7

the seafloor spreads at a rate of aproximatly 10 cm per year. if you were to collect data on the spread of the seafloor each wee

k, which unit whoul you use to record your data. Explain?
Mathematics
1 answer:
Natali [406]3 years ago
5 0

Answer:

2,000 micrometers

Step-by-step explanation:

Well 10 cm in 50 or so weeks is 0.2 cm per week so you could use mm  

Then the measuremet would be about 2 mm per week.  

If the measurement is extremely accurate you could consider using micrometres.  

There are 1000 micrmetres in 1mm so in 1 week 2,000 micrometres

Hope this helps! <3 Jacoby


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Option C:

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Solution:

Given expression:

$\frac{\left(5 g^{4}+5 g^{3}-17 g^{2}+6 g\right)-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right)}{g+2}

To find which expression is equal to the given expression.

$\frac{\left(5 g^{4}+5 g^{3}-17 g^{2}+6 g\right)-\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right)}{g+2}

Expand the term -\left(3 g^{4}+6 g^{3}-7 g^{2}-12\right):-3 g^{4}-6 g^{3}+7 g^{2}+12

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Arrange the like terms together.

         $=\frac{5 g^{4}- 3 g^{4}+5 g^{3}-6 g^{3}-17 g^{2}+7 g^{2}+6 g+12}{g+2}

         $=\frac{2 g^{4}- g^{3}-10 g^{2}+6 g+12}{g+2}

Factor the numerator 2 g^{4}-g^{3}-10 g^{2}+6 g+12=(g+2)\left(2 g^{3}-5 g^{2}+6\right)

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Cancel the common factor g + 2, we get

         =2 g^{3}-5 g^{2}+6

Hence option C is the correct answer.

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