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3 years ago

Compare 0.4____6.22. Use <, >, or =

Mathematics

2 answers:

NNADVOKAT [17]3 years ago

7 0

In this problem your answer would be < the gator likes larger numbers

STatiana [176]3 years ago

3 0

0.4 < 6.22 is correct and so since this is not false, there are infinitely many answers.

Answer: A) because we know 6.22 is not lesser than and we know its not equal, so its greater than

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Step-by-step explanation:

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3 years ago

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2 years ago

Find the quotient. 12a^3p^4 ÷ -2a^2p

777dan777 [17]

Answer:

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Step-by-step explanation:

We have to find the quotients of

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3 years ago

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$

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3 years ago