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2 years ago

If f(x) is an exponential function where f(2)=23 and f(3)=58, then find the value of f(6), to the nearest hundredth.

Mathematics

1 answer:

Stels [109]2 years ago

8 0

The exponential function is given as 3.68 × (2.52)ˣ. Then the value of function at x = 6 will be 942.43.

<h3>What is an exponent?</h3>

The exponent denotes that the base will rise to a specific level of strength. The base is X, and the power is n.

If f(x) is an exponential function where f(2) = 23 and f(3) = 58.

Then the exponential function will be

f(x) = abˣ

If f(2) = 23, then we have

23 = ab² ...1

If f(3) = 58, then we have

58 = ab³ ...2

Then from equations 1 and 2, we have

a = 3.62 and b = 2.52

Then the function will be

f(x) = 3.68 × (2.52)ˣ

Then the value of f(6) will be

f(6) = 3.68 × (2.52)⁶

f(6) = 3.68 × 256.1

f(6) = 942.43

More about the exponent link is given below.

brainly.com/question/5497425

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I hope it helps.

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3 years ago

The circle graph shows how a family spends its annual income. If $26,450 is used for Auto and Entertainment combined, what is th

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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)

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: $F=0.055 L\s$

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