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2 years ago

Rewrite the following as three reduced terms using fractional coefficients and negative exponents if needed:

Mathematics

1 answer:

Elan Coil [88]2 years ago

5 0

Answer:

1/2x^2 -5/6 +x^(-2)

Step-by-step explanation:

The sum can be separated into its components, and each of those simplified to the desired form.

<h3>Rewrite</h3>

$\dfrac{3x^4-5x^2+6}{6x^2}=\dfrac{3x^4}{6x^2}+\dfrac{-5x^2}{6x^2}+\dfrac{6}{6x^2}\\\\=\boxed{\dfrac{1}{2}x^2-\dfrac{5}{6}+x^{-2}}$

<em>Additional comment</em>

The rules of exponents tell you ...

(a^b)/(a^c) = a^(b-c)

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3 years ago

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2 years ago

If john run 2.5 miles everyday, how many miles will he run in a month of 30?

Lapatulllka [165]

Answer:

2,250 miles per day.

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John runs 2.5 miles per day. In a period of 30 months, you can averagely take 30 days in each month.

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900 x 2.5 = 2,250 miles per day.

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3 years ago

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d

nikitadnepr [17]

Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

$P(t)=Ce^{kt}$

c) 977,435,644 cells

d) 2.033 billions cells per hour.

e) 2.81 hours.

Step-by-step explanation:

We have a model of exponential growth.

We know that the population duplicates every 20 minutes (t=0.33).

The initial population is P(t=0)=58.

The exponential growth model can be written as:

$P(t)=Ce^{kt}$

For t=0, we have:

$P(0)=Ce^0=C=58$

If we use the duplication time, we have:

$P(t+0.33)=2P(t)\\\\58e^{k(t+0.33)}=2\cdot58e^{kt}\\\\e^{0.33k}=2\\\\0.33k=ln(2)\\\\k=ln(2)/0.33=2.08$

Then, we have the model as:

$P(t)=58e^{2.08t}$

The relative growth rate (RGR) is defined, if P is the population and t the time, as:

$RGR=\dfrac{1}{P}\dfrac{dP}{dt}=k$

In this case, the RGR is k=2.08 1/h.

After 8 hours, we will have:

$P(8)=58e^{2.08\cdot8}=58e^{16.64}=58\cdot 16,852,338= 977,435,644$

The rate of growth can be calculated as dP/dt and is:

$dP/dt=58[2.08\cdot e^{2.08t}]=120.64e^2.08t=2.08P(t)$

For t=8, the rate of growth is:

$dP/dt(8)=2.08P(8)=2.08\cdot 977,435,644 = 2,033,066,140$

(2.033 billions cells per hour).

We can calculate when the population will reach 20,000 cells as:

$P(t)=20,000\\\\58e^{2.08t}=20,000\\\\e^{2.08t}=20,000/58\approx344.827\\\\2.08t=ln(344.827)\approx5.843\\\\t=5.843/2.08\approx2.81$

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3 years ago