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dem82 [27]
3 years ago
11

A scientist measured the width of a square flake of gold to be7.3 x 10 -4mm what is the area of the flake

Mathematics
1 answer:
irakobra [83]3 years ago
7 0

The area of square flake of gold is 53.29 \times 10^{-8} \mathrm{mm}

<h3><u>Solution:</u></h3>

Given that width of square flake of gold is 7.3 \times 10^{-4} millimeter

To find : Area of square flake

Since gold flake is in shape of square, we can use area of square formula

<em><u>The formula for area of square is given as:</u></em>

area = s^2

where "s" is the length of one side

\begin{array}{l}{\text { Area of a square flake }=\left(7.3 \times 10^{-4}\right) \times\left(7.3 \times 10^{-4}\right)} \\\\ {\text { Area of a square flake }=53.29 \times 10^{8}}\end{array}

Hence , the calculated area of flake is 53.29 \times 10^{-8} \mathrm{mm}

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6 0
3 years ago
F(z) = (z - 5)(2x + 7) (73 = 3) has zeros at x = -3.5, x=3/7 , x=5 What is the sign of f on the interval 3/7 &lt; x &lt;5?
Gwar [14]
<h3>Answer:     Negative</h3>

======================================================

Explanation:

3/7 = 0.42857 approximately

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Plug x = 2 into the f(x) function

f(x) = (x - 5)(2x + 7)(7x-3)

f(2) = (2 - 5)(2*2 + 7)(7*2-3)

f(2) = (2 - 5)(4 + 7)(14-3)

f(2) = (-3)(11)(11)

f(2) = -363

The actual result doesn't matter. All we're after is whether the result is positive or negative. We see the result is negative. This means f(x) is negative when 3/7 < x < 5. The f(x) curve is below the x axis on this interval.

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3 years ago
Problem PageQuestion The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate paramet
asambeis [7]

Answer:

<em>t = 1.51</em>

Step-by-step explanation:

<u>Exponential Model</u>

The exponential model is often used to simulate the behavior of a magnitude that either grow or decay in proportion to the existing amount of that magnitude.

The model can be expressed as

M=M_oe^{kt}

In this case, Mo is the initial mass of the radioactive substance and k is a constant which value is positive if the mass is growing or negative if the mass is decaying.

The value of k is not precisely given in the question, we are assuming k=-0.2

The model is now

M=M_oe^{-0.2t}

We are required to compute the time it takes the mass to reach one-half of its initial value:

\displaystyle \frac{M_o}{2}=M_oe^{-0.2t}

Simplifying

\displaystyle \frac{1}{2}=e^{-0.2t}

Taking logarithms

\displaystyle ln\frac{1}{2}=ln(e^{-0.2t})=-0.2t

Solving for t

\displaystyle t=-\frac{ln\frac{1}{2}}{0.2}=1.51

6 0
3 years ago
What is the solution to the system of equations below?
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Answer: Infinite solutions.

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Plug the value of y into the second equation.

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Add 12 on both sides.

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Add 4x on both sides.

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6 0
3 years ago
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I hope this helps:)

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3 years ago
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