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1 year ago

Use the function below to find f(1). f9t0=4

Mathematics

1 answer:

ExtremeBDS [4]1 year ago

6 0

The value of the F(1) is 1/2 option (C) is correct after plugging the value of t = 1 in the provided function.

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

The question is incomplete.

The complete question is attached in the picture please refer to the picture.

We have a function:

$\rm F(t) = 4\dfrac{1}{2^{3t}}$

Plug t = 1

$\rm F(1) = 4\dfrac{1}{2^{3\times1}}$

After solving:

F(1) = 1/2

Thus, the value of the F(1) is 1/2 option (C) is correct after plugging the value of t = 1 in the provided function.

Learn more about the function here:

brainly.com/question/5245372

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3 years ago

Find the mistake made in the steps to solve the equation below.

IgorC [24]

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Step-by-step explanation:

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Store A sells 12 juice bottles for $4 and store B sells 18 juice bottle for $6. Are the rates equivalent? explain your reasoning

mamaluj [8]

Answer:

Yes!

The rate can be expressed by the ratio 3:1

For every 3 bottles, its $1

For every 6 bottles, its $2

For every 12 bottles its $4

and for every 18 bottles its $6

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Step-by-step explanation:

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3 years ago

A particle moves along a line with acceleration a (t) = -1/(t+2)2 ft/sec2. Find the distance traveled by the particle during the

amm1812

Answer:

$s(t)=(ln|7|+ln|2|)\,ft$

Step-by-step explanation:

Acceleration is second derivative of distance and are related as:

$a(t)=\frac{d^2s}{dt^2}\\\\\frac{d^2s}{dt^2}=\frac{-1}{(t+2)^2}\\$

Integrating both sides w.r.to t

$v(t)=\frac{ds}{dt}=\frac{1}{t+2} +C\\$

Using initial value

$v(0)=\frac{1}{2}\\\\\frac{1}{2}=\frac{1}{0+2} +C\\\\C=0\\\\\frac{ds}{dt}=\frac{1}{t+2}$

We have to calculate the distance covered in time interval [0,5], so:

$\int\limits^5_0 \frac{ds}{dt}=\int\limits^5_0 {\frac{1}{t+2}} \, dt\\\\s(t)=[ln|t+2|]^5_0\\\\s(t)=ln|5+2|+ln|0+2|\\\\s(t)=(ln|7|+ln|2|)\,ft$

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3 years ago