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3 years ago

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What is the sum of the arithmetic series 18 sigma t-1 (5t-4)?

2 answers:

just olya [345]3 years ago

3 0

Answer:

783

Step-by-step explanation:

We have to find the sum of Arithmetic series from t = 1 to t = 18 represented by (5t - 4)

The formula to find the sum of an Arithmetic series when first and the last term is known is:

$S_{n}=\frac{n}{2}(a_{1}+a_{n})$

Here,

n = Total number of terms = 18

$a_{1}$ = First Term = 5(1) - 4 = 1

$a_{18}$ = 18th Term = 5(18) - 4 = 86

Using the values in the above formula, we get:

$S_{18}=\frac{18}{2}(1+86)\\\\ S_{18}=9(87)\\\\ S_{18}=783$

Thus, the sum of 18 terms of the given Arithmetic Series is 783.

Tomtit [17]3 years ago

3 0

Question :

What is the sum of the arithmetic series 18 sigma t-1 (5t-4)?

Answer & Step-by-step explanation:

We have to find the sum of Arithmetic series from t = 1 to t = 18 represented by (5t - 4)

The formula to find the sum of an Arithmetic series when first and the last term is known is:

$S_{n}=\frac{n}{2}(a_{1}+a_{n})$

n = Total number of terms = 18

$a_{1}$ = First Term = 5(1) - 4 = 1

$a_{18}$ = 18th Term = 5(18) - 4 = 86

Using the values in the above formula, we get:

$S_{18}=\frac{18}{2}(1+86)\\S_{18}=9(87)\\ S_{18}=783$

Thus, the sum of 18 terms of the given Arithmetic Series is 783.

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A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

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and flows out at a rate

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Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

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