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Reika [66]
3 years ago
10

Find the sum of the first 8 terms of the series. 1/2 + 1 + 2 + 4

Mathematics
2 answers:
Fynjy0 [20]3 years ago
5 0

Answer:

The sum of 8 term of the series is \dfrac{255}{2} or 127.5

Step-by-step explanation:

Given: The series \frac{1}{2} + 1 +2 +4 + ...

We need to find the sum of 8 term

Common ratio, r =\dfrac{2}{1}= 2

First term, a=\dfrac{1}{2}

n=8

Formula:

S_n=\dfrac{a(r^n-1)}{(r-1)}

Substitute the value of a, r and n into formula

Sum of 8th term of the sequence

S_8=\dfrac{1/2(2^8-1)}{2-1}

S_8=\dfrac{255}{2}

Hence, The sum of 8 term of the series is \dfrac{255}{2} or 127.5

Roman55 [17]3 years ago
4 0

\bf \cfrac{1}{2}~~,~~\stackrel{2\cdot \frac{1}{2}}{1}~~,~~\stackrel{2\cdot 1}{2}~~,~~\stackrel{2\cdot 2}{4}~~,...

so as you can see the common ratio is 2, and the first term is 1/2,

\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad  \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ r=2\\ a_1=\frac{1}{2}\\ n=8 \end{cases} \\\\\\ S_8=\cfrac{1}{2}\left( \cfrac{1-2^8}{1-2} \right)\implies S_8=\cfrac{1}{2}\left( \cfrac{-255}{-1} \right)\implies S_8=\cfrac{255}{2}

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7 0
3 years ago
The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
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Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

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3 years ago
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Gemiola [76]

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3 years ago
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