Suppose that dP/dt = 0.19P(t) represents a mathematical model for the growth of a certain cell culture, where P(t) is the size of the culture (measured in millions of cells) at time t > 0 (measured in hours). How fast is the culture growing at the time t when the size of the culture reaches 2 million cells?

Answers

Answer 1
Answer:

Answer: 380000 cells/hour

Step-by-step explanation:

Given that dP/dt = 0.19P(t)

where

P(t) is the size of the culture (measured in millions of cells) at time t > 0 (measured in hours).

The formula above represents a mathematical model for the growth of a certain cell culture. In essence, it represents the time rate of the growth of the cell culture, that is how fast the cell culture is growing.

Therefore, when P = 2 million cells:

dP/dt = 0.19 * 2000000 = 380000 cells/hour

Hence, the cell culture is growing at 380000 cells per hour.

Answer 2
Answer:

Final answer:

The cell culture is growing at a rate of 0.38 million cells per hour when the size of the culture reaches 2 million cells. This is based on the given differential equation dP/dt = 0.19P(t).

Explanation:

This problem is associated with the concept of differential equations, particularly exponential growth. In this scenario, the rate of change of P(t), the size of the cell culture, is given by the equation dP/dt = 0.19P(t). This can be interpreted as the culture is growing at 19% per unit of time.

To determine how fast the culture is growing when P(t) equals 2 million cells, we need to substitute 2 for P(t) in the given equation: dP/dt = 0.19*2. This calculation returns a value of 0.38 million cells per hour. Therefore, when the size of the culture reaches 2 million cells, it is growing at a rate of 0.38 million cells per hour.

Learn more about Exponential Growth

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The sum of three consecutive odd numbers is 69. What is the smallest of the three numbers?

Answers

Answer:

21

Step-by-step explanation:

Let the smallest of the numbers be N

The other two numbers (consecutive) would be written as (n + 2), (n + 4)

Expressing these as an equation gives : (n) + (n+2) + (n+4) = 69

opening the bracket and collecting like terms, we have:

n+n+n+2+4=69

3n + 6 = 69

3n = 69 - 6

3n = 63

Dividing both sides by 3 or making n the subject formular, we get:

n = 63/3

n = 21.

Note, the other numbers are: 21, 23, and 25

They are all odd numbers

Their sum equals to 69

Eight times the difference of y and nine

Answers

Eight times the difference of y and nine will be 8(y - 9).

It should be noted that eight times the difference of y and nine simply means that one has to subtract 9 from y and then multiply the difference by 8.

Therefore, eight times the difference of y and nine will be 8(y - 9).

In conclusion, the correct option is 8(y - 9).

Read related link on:

brainly.com/question/16081696

Answer:

(y-9)8

Step-by-step explanation:

you first solve 8-9, and then multiply is by 8.

Jack has a summer job caddying at a golf course. Jack earns $120a week. By rounding his weekly income to the nearest hundred
dollars, find a reasonable estimate for his total income during the
12 week summer.

A) $700
B) $800
C) $1200
D) $1700

Answers

1200 as you would round it down to 100 a week so you would do 100x12

6 apple trees fir every 4 pears trees How many apple trees would there be if there were 42 pear trees

Answers

Answer: 63 apple trees
Explanation: divide 42 by 4 to get that amount of groups (10.5) then multiply that number by 6 so (10.5 • 6) and that should give the apple tree total.

Please Help!!In 2010, Martin paid $5,518.00 in Social Security tax. The SS tax rate was 6.2%. What was Martin's taxable income in 2010?

Answers

Answer:

$890 or it's $34211.6

Step-by-step explanation:

you divide 5518 by 6.2% or you multiply them

Solve this differential Equation by using power series
y''-x^2y=o

Answers

We're looking for a solution

y=\displaystyle\sum_(n=0)^\infty a_nx^n

which has second derivative

y''=\displaystyle\sum_(n=2)^\infty n(n-1)a_nx^(n-2)=\sum_(n=0)^\infty(n+2)(n+1)a_(n+2)x^n

Substituting these into the ODE gives

\displaystyle\sum_(n=0)^\infty(n+2)(n+1)a_(n+2)x^n-\sum_(n=0)^\infty a_nx^(n+2)=0

\displaystyle\sum_(n=0)^\infty(n+2)(n+1)a_(n+2)x^n-\sum_(n=2)^\infty a_(n-2)x^n=0

\displaystyle2a_2+6a_3x+\sum_(n=2)^\infty(n+2)(n+1)a_(n+2)x^n-\sum_(n=2)^\infty a_(n-2)x^n=0

\displaystyle2a_2+6a_3x+\sum_(n=2)^\infty\bigg((n+2)(n+1)a_(n+2)-a_(n-2)\bigg)x^n=0

Right away we see a_2=a_3=0, and the coefficients are given according to the recurrence

\begin{cases}a_0=y(0)\na_1=y'(0)\na_2=0\na_3=0\nn(n-1)a_n=a_(n-4)&\text{for }n\ge4\end{cases}

There's a dependency between terms in the sequence that are 4 indices apart, so we consider 4 different cases.

  • If n=4k, where k\ge0 is an integer, then

k=0\implies n=0\implies a_0=a_0

k=1\implies n=4\implies a_4=(a_0)/(4\cdot3)=\frac2{4!}a_0

k=2\implies n=8\implies a_8=(a_4)/(8\cdot7)=(6\cdot5\cdot2)/(8!)a_0

k=3\implies n=12\implies a_(12)=(a_8)/(12\cdot11)=(10\cdot9\cdot6\cdot5\cdot2)/(12!)a_0

and so on, with the general pattern

a_(4k)=(a_0)/((4k)!)\displaystyle\prod_(i=1)^k(4i-2)(4i-3)

  • If n=4k+1, then

k=0\implies n=1\implies a_1=a_1

k=1\implies n=5\implies a_5=(a_1)/(5\cdot4)=(3\cdot2)/(5!)a_1

k=2\implies n=9\implies a_9=(a_5)/(9\cdot8)=(7\cdot6\cdot3\cdot2)/(9!)a_1

k=3\implies n=13\implies a_(13)=(a_9)/(13\cdot12)=(11\cdot10\cdot7\cdot6\cdot3\cdot2)/(13!)a_1

and so on, with

a_(4k+1)=(a_1)/((4k+1)!)\displaystyle\prod_(i=1)^k(4i-1)(4i-2)

  • If n=4k+2 or n=4k+3, then

a_2=0\implies a_6=a_(10)=\cdots=a_(4k+2)=0

a_3=0\implies a_7=a_(11)=\cdots=a_(4k+3)=0

Then the solution to this ODE is

\boxed{y(x)=\displaystyle\sum_(k=0)^\infty a_(4k)x^(4k)+\sum_(k=0)^\infty a_(4k+1)x^(4k+1)}