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ra1l [238]
2 years ago
6

How many solutions does the following have. Y=3x-2 and 2x+5y=7

Mathematics
1 answer:
bezimeni [28]2 years ago
3 0
One, the intersection point is (1,1)
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Solve log2(x) – log2(3) = 2.
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Step-by-step explanation:

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What is the volume of a hemisphere with a radius of 2.3 m, rounded to the nearest tenth of a cubic meter?
ANTONII [103]

Answer:25.5m^3

Step-by-step explanation:

radius(r)=2.3m

π=3.14

volume of hemisphere=2/3 x π x r^3

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Volume=2/3 x 3.14 x 2.3 x 2.3 x 2.3

volume=(2x3.14x2.3x2.3x2.3) ➗ 3

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3 0
3 years ago
EXPLAIN
zvonat [6]

Answer:

50,000? This is my answer sorry if wrong

Step-by-step explanation:

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2 years ago
Dan sold 40 concert tickets in 5 days. Each day he sold 3 tickets more than the previous day the number of tickets he sold on th
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5 0
2 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}

6 0
3 years ago
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