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Korolek [52]
2 years ago
8

Help for question 1 and 2​

Mathematics
1 answer:
AnnyKZ [126]2 years ago
4 0

Answer:

<em>(1). </em>5^{4}<em> × </em>13^{7}<em> ; (2). 3 - 5x</em>

Step-by-step explanation:

1). [( - 5 )³ · ( - 5 )] ( 13^{-2} · 13^{9} ) = (-5)^{3+1} × (13)^{-2+9} = (-5)^{4} × 13^{7} = 5^{4} × 13^{7}

2). - 25x² ÷ 5x + 15x ÷ 5x = <em>3 - 5x</em>

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2 years ago
a 34 gram sample of a substance that's used to treat thyroid disorders has a k-value of 0.137. find the substance's half-life, i
Ira Lisetskai [31]

Answer:

5.1 days

Step-by-step explanation:

Given in the question,

initial amount of substance = 34 grams

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To find the half life of this substance we will use following formula

N(0)/2 = N(0)e^{-kt}

here N(0) is initial amount of substance

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Plug values in the formula

34 /2 = 34e^{-0.137t}

1/2 = e^{-0.137t}

Take logarithm on both sides

ln(1/2) = ln( e^{-0.137t})

ln(1/2) = -0.137t

t = ln(1/2) / -0.137

t = 5.059

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3 0
3 years ago
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5 0
3 years ago
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