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polet [3.4K]
2 years ago
9

Need help with R3, R4, S5​

Mathematics
1 answer:
marissa [1.9K]2 years ago
7 0

Answer:

R3 Theorem: If parallel lines are cut by a transversal, then alternate interior angles are congruent.

R4 Theorem: Congruence of segments is reflexive.

S5 ΔABC ≅ ΔCDA

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Technetium-99m is used as a radioactive tracer for certain medical tests. It has a half-life of 1 day. Consider the function T w
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Answer:

The expression that represents the number of days until only 10% remains is T((d) 10 %) =100×(\frac{1}{2} )^{3.322}.

Step-by-step explanation:

The equation for half life is of the form

A = A₀×(\frac{1}{2} )^{\frac{t}{h} }.........................................................................(1)

Where

A = Final amount

A₀ = Initial amount

t = Time

h = Half life

For the equation T(d) = 100×2⁽⁻²⁾....................................(2)

We have by comparison with the equation for half life

2 ≡ \frac{t}{h}  and and the equation (2) can be written as

Percentage remaining after 2 half lives is

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{2 }

However if the half life of Technetium-99m is 6 hours then we have for one day

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{2 *2}

Therefore an expression that represents the number of days until only 10% remains is

\frac{A}{A_0} ×100=100×(\frac{1}{2} )^{\frac{d}{h}  } = 10 %

(\frac{1}{2} )^{\frac{d}{h}  } =\frac{1}{10}

= ㏑(\frac{1}{2} )^{\frac{d}{h}  }  = ㏑(\frac{1}{10})

= \frac{d}{h}×㏑(\frac{1}{2} ) = ㏑(\frac{1}{10})

\frac{d}{h} = \frac{ln(\frac{1}{10}) }{ln(\frac{1}{2} )} = 3.322

Therefore the expression for the number of days 10 % of Technetium-99m will be remaining is

T((d) 10 %) =100×(\frac{1}{2} )^{3.322}

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