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user100 [1]
3 years ago
6

A radio active isotope decays according to the exponential decay equation where t is in days. Round to the thousandths place. Th

e half life is the solution (t) of the equation: a/2=ae^-2.360t
Mathematics
1 answer:
Whitepunk [10]3 years ago
8 0
To calculate the value of the half life we proceed as follows;
a/2=ae^(-2.360t)
the a(s) will cancel and we shall remain with:
1/2=e^-2.360t
introducing the natural logs we get:
ln (1/2)=ln e^(-2.360t)
ln(1/2)=-2.360t
hence;
t=(ln 0.5)/(-2.360)
t=0.294
the answer is t=0.294
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Cassandra earns $12.98 an hour working at Key Food. How much money will she
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3 years ago
Select the correct answer from each drop-down menu. In the figure, . Angles are congruent . ∠GAC ≅ ∠AFE because they are corresp
Hoochie [10]

Answer:

∠GAC ≅ ∠HFD by the Property of Congruence.

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I'm going to be honest the question is a little confusing cause of the beginning, but if you're looking for which angles are actually congruent it's ∠GAC ≅ ∠HFD

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2 years ago
an aquarium has a ratio of 5 rays for every 3 sharks. What percent of this grouping of animals are rays
sashaice [31]

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62.5%

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6 0
3 years ago
The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.
tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}

so, it passes through the midpoint of AB,

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}

4 0
3 years ago
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