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vagabundo [1.1K]
2 years ago
9

What is the solution to the equation below?Round your answer to two decimal places.6+7*log2x=21

Mathematics
2 answers:
Kipish [7]2 years ago
5 0
Are you looking for x
Sloan [31]2 years ago
4 0

Answer:

Solution of 6+7 log 2x=21 is x = 4.26

Step-by-step explanation:

We need to find solution of 6+7 log 2x=21

Solving

          6+7 log 2x=21

         7 log 2x=21 - 6

          7 log 2x=15

          log 2x= 2.14

          2x = 8.52

            x = 4.26

Solution of 6+7 log 2x=21 is x = 4.26

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Next you want to find the area of the 120 degree section of that circle.
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3 years ago
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katrin2010 [14]

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Step-by-step explanation:

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6 0
3 years ago
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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)
Lemur [1.5K]

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

4 0
2 years ago
What is the value of the expression 4⁴?​
Ghella [55]

Answer:

256 is the answer

Step-by-step explanation:

4×4×4×4=256.

3 0
3 years ago
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