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2 years ago

How many centigrams are in 6 hectograms? Use the metric table to help answer the question.

Mathematics

2 answers:

Marta_Voda [28]2 years ago

4 0

There are 60000 centigrams in 6 hectograms.

Over [174]2 years ago

4 0

Answer:

There are 60000 centigrams in 6 hectograms

Step-by-step explanation:

We know that 1 hectogram=10000 centigrams

Multiplying both sides by 6

⇒ 6 hectograms=6×10000 centigrams

⇒ 6 hectograms=60000 centigrams

Hence, there are 60000 centigrams in 6 hectograms

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After the mill in a small town closed down in 1970, the population of that town started decreasing according to the law of expon

wlad13 [49]

Answer:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

Step-by-step explanation:

The natural growth and decay model is given by:

$\frac{dP}{dt}=kP$ (1)

Where P represent the population and t the time in years since 1970.

If we integrate both sides from equation (1) we got:

$\int \frac{dP}{P} =\int kdt$

$ln|P| =kt +c$

And if we apply exponentials on both sides we got:

$P= e^{kt} e^k$

And we can assume $e^k = P_o$

And we have this model:

$P(t) = P_o e^{kt}$

And for this case we want to find $P_o$

By 1990 we have t=20 years since 1970 and we have this equation:

$143000 = P_o e^{20k}$

And we can solve for $P_o$ like this:

$P_o = \frac{143000}{e^{20k}}$ (1)

By 2019 we have 49 years since 1970 the equation is given by:

$98000 = P_o e^{49k}$ (2)

And replacing $P_o$ from equation (1) we got:

$98000=\frac{143000}{e^{20k}} e^{49k} =143000 e^{29k}$

We can divide both sides by 143000 we got:

$\frac{98000}{143000} =0.685 = e^{29k}$

And if we apply ln on both sides we got:

$ln(0.685) = 29k$

And then k =-0.01303024661[/tex]

And replacing into equation (1) we got:

$P_o = \frac{143000}{e^{-20*0.01303024661}}=110193.69$

And we can round this to the nearest up integer and we got 110194.

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Step by step explanation on how to solve -22-(6)(-9)/3

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Answer:

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Step-by-step explanation:

-22-(6)(-9)/3

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