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Katyanochek1 [597]

2 years ago

14

need help asap!

1 answer:

Sonbull [250]2 years ago

4 0

Answer:

They are simplified in hopes of finding like terms so we can add or subtract them.

When multiplying or dividing we can multiply or divide the numbers under the radicals before simplifying

Step-by-step explanation:

√8 and √32 are not like terms, so we can't simply add or subtract them. We first want to reduce them down as much as possible...

√8 + √32

√8 = (√4)(√2) = 2√2

√32 = (√16)(√2) = 4√2

So √8 + √32 is the same as 2√2 + 4√2

These are now like terms since they share a common number under the radical sign.

2√2 + 4√2 = 6√2

Multiplying: (√2)(√2) = √4 = 2

Dividing: (√16)/(√4) = √4 = 2

or (√16)/(√4) = 4/2 = 2

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Identify which lines are parallel y=(1/2)× 4

mars1129 [50]

Your question is a bit unclear, but if you mean which lines are parallel to this one then I have an answer.

1/2*4=2 so this line would be y=2 which means it is a horizontal line. Therefore, any other horizontal line would be parallel to it. So just pick a value for Y to equal and that will be your answer.
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5 0

3 years ago

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Ray Of Light [21]

Answer is 3
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4 0

3 years ago

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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

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SOVA2 [1]

7h + 43 = 64

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5 0

2 years ago

Solve the system using substitution.

Alexeev081 [22]

Answer:

Option B that is (2,6) is correct

Explanation:

We have been given with system of equations

$X+Y=8$ and $Y=3X$

Here, we will substitute the value of $Y=3X$ in $X+Y=8$

We will get $X+3X=8$

Further simplification we will get to $4X=8$

Hence, the value of $X=2$

And now, substituting the value $X=2$ in $Y=3X$ we will get $Y=6$ .

Therefore, option B is correct.

4 0

3 years ago

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