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Katyanochek1 [597]
3 years ago
5

If you start with 85 milligrams of Chromium 51, used to track red blood cells, which

Mathematics
1 answer:
zysi [14]3 years ago
8 0

About 92 days are taken for 90 % of the material to <em>decay</em>.

The mass of radioisotopes (m), measured in milligrams, decreases exponentially in time (t), measured in days. The model that represents such decrease is described below:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} } (1)

Where:

  • m_{o} - Initial mass, in milligrams.
  • m(t) - Current mass, in milligrams.
  • \tau - Time constant, in days.

In addition, the time constant is defined in terms of half-life (t_{1/2}), in days:

\tau = \frac{t_{1/2}}{\ln 2} (2)

If we know that m_{o} = 85\,mg, t_{1/2} = 27.7\,d and m(t) = 8.5\,mg, then the time required for decaying is:

\tau = \frac{t_{1/2}}{\ln 2}

\tau = \frac{27.7\,d}{\ln 2}

\tau \approx 39.963\,d

t = -\tau \cdot \ln \frac{m(t)}{m_{o}}

t = -(39.963\,d)\cdot \ln \frac{8.5\,mg}{85\,mg}

t\approx 92.018\,d

About 92 days are taken for 90 % of the material to <em>decay</em>.

We kindly invite to check this question on half-life: brainly.com/question/24710827

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Two hoses are filling a pool the first hose fills at a rate of x gallons per minute the second hose fills at a rate of 15 gallon
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Answer:

B. (0, 5]∪(15,30] only (15,30] contains viable rates for the hoses.

Step-by-step explanation:

The question is incomplete. Find the complete question in the comment section.

For us to meet the pool maintenance company's schedule, the pool needs to fill at a combined

rate of at least 10 gallons per minute. If the inequality represents the combined rates of the hoses is 1/x+1/x-15≥10 we are to find all solutions to the inequality and identifies which interval(s) contain viable filling rates for the  hoses. On simplifying the equation;

\frac{1}{x} + \frac{1}{x-15} \geq \frac{1}{10}\\\\ find\ the \  LCM \ of \ the function \ on \ the \ LHS\\\\\frac{x-15+x}{x(x-15)} \geq \frac{1}{10}\\\\\frac{2x-15}{x(x-15)} \geq  \frac{1}{10}\\\\10(2x-15)\geq x(x-15)\\\\20x-150\geq x^2-15x\\\\collect \ like \ terms\\-x^2+20x+15x - 150\geq 0\\

-x^2+35x-150 \geq 0\\\\multipply \ through \ by \ minus\\x^2-35x+150 \leq  0\\\\(x^2-5x)-(30x+150) \leq  0\\\\x(x-5)-30(x-5) \leq 0\\\\

(x-5)(x-30) \leq 0\\\\x-5 \leq 0 and x - 30 \leq 0\\\\x \leq  5 \ and \ x \leq 30

The interval contains all viable rate are values of x that are less than 30. The range of interval is (0, 5]∪(15,30]. Since the pool needs to fill at a combined  rate of <em>at least 10 gallons per minute</em> for the pool to meet the company's schedule, <em>this means that the range of value of gallon must be more than 10, hence (15, 30] is the interval that contains the viable rates for the hoses.</em>

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