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Ksju [112]
3 years ago
11

Rosita had 17 subtraction problems and 28 multiplication problems. If she ca finish five problems in 20 minutes, how long will i

t take her to finish all the problems?
Mathematics
2 answers:
lana [24]3 years ago
8 0

Step-by-step explanation:

total problems = 17+28=45

If she ca finish five problems in 20 minutes,

so...

for 45 problems

she take 45/5*20=180 minute = 3hour

tamaranim1 [39]3 years ago
5 0

Answer:

It would be 3 hours

Step-by-step explanation: you add 17+28 and you get 45, then you divide 45by 5 which gets you 9 and then you add 20 together 9 times

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jok3333 [9.3K]

Answer: The percent gain was 25%.

Step-by-step explanation:

Given, Cost price of lot = $1,200

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3 years ago
Write the first four terms in the following sequences. A(n+1)=12 A(n) for n≥1 and A(1)=4 .
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3 years ago
PLEASE HELP ASAP!! SPAM ANSWERS WILL BE DELETED
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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
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\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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