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2 years ago

Find the next three terms for the following arithmetic sequence: -3,-1, 1,

Mathematics

2 answers:

Alecsey [184]2 years ago

5 0

Answer:

3 and 5

Step-by-step explanation:

The common difference d in an arithmetic sequence is

d = a₂ - a₁ = - 1 - (- 3) = - 1 + 3 = 2

To obtain a term in the sequence add d = 2 to the previous term, that is

a₄ = a₃ + 2 = 1 + 2 = 3

a₅ = a₄ + 2 = 3 + 2 = 5

Tems11 [23]2 years ago

3 0

Answer:

3, 5

Step-by-step explanation:

It is counting up by +2

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Which of the following shows the coordinates of A (6, 12) after reflection over the y-axis?

lorasvet [3.4K]

Answer:

(- 6, 12 )

Step-by-step explanation:

Under a reflection in the y- axis

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A (6, 12 ) → A' (- 6, 12 )

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2 years ago

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2 years ago

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2 years ago

A culture started with 6,000 bacteria. After 4

Oksanka [162]

19539 bacteria will be present after 18 hours

Solution:

Initial value of bacteria N = 6000

Value after 4 hours $N_o$ = 7800

The standard exponential equation is given as:

$N=N_{o} E^{-k t}$

where

N is amount after time t

No is the initial amount

k is the constant rate of growth

t is time

Plugging in the values in formula we get,

$7800 = 6000E^{-4k}$

Solving for "k" we get,

$\frac{13}{10}=E^{-4k}$

Taking "ln" on both sides, we get

$ln\frac{13}{10} = -4k$

$\frac{ln\frac{13}{10}}{4} = -k$

On solving for ln, we get k = -0.0656

The equation becomes, $N = 6000E^{0.0656t}$

Now put "t" = 18,

$N = 6000E^{0.0656 \times 18}\\\\N = 6000 \times 3.256 = 19539$

Hence the bacteria present after 18 hours is 19539

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2 years ago

Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.

Veseljchak [2.6K]

This equation is separable, as

$\dfrac{\mathrm dy}{\mathrm dx}=y(y-2)e^x\implies\dfrac{\mathrm dy}{y(y-2)}=e^x\,\mathrm dx$

Integrate both sides; on the left, expand the fraction as

$\dfrac1{y(y-2)}=\dfrac12\left(\dfrac1{y-2}-\dfrac1y\right)$

Then

$\displaystyle\int\frac{\mathrm dy}{y(y-2)}=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C$

$\implies\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x+C$

Since $y(0)=1$ , we get

$\dfrac12\ln\left|\dfrac{1-2}1\right|=e^0+C\implies C=-1$

so that the particular solution is

$\dfrac12\ln\left|\dfrac{y-2}y\right|=e^x-1\implies\boxed{y=\dfrac2{1-e^{2e^x-2}}}$

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2 years ago