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andrew-mc [135]
3 years ago
11

What is the volume of a sphere with the radius of 5 ft?

Mathematics
1 answer:
konstantin123 [22]3 years ago
6 0
The answer is 523.6ft3



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2x−3, find h(6)h(6).
denis-greek [22]

Answer:

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3 0
3 years ago
Read 2 more answers
True or false? Do the following set of ordered pairs represent a function {(8,1),(0,-1),(9,4),(-6,-7),(8,9)}
Anika [276]
Hi there!

Unfortunately, the set of ordered pairs does NOT represent a function! This is because there are two of the same x values. In a function, there is one input for every output. In this case, there are two of the same inputs. However, there can be two of the same outputs. 
Just to clarify - Not a function

Hope this helps!! :)
If there's anything else that I can help you with, please let me know! 
6 0
3 years ago
What is the numerical coefficient in y=−6x−3
liq [111]
The numerical coefficient in y= -6x-3 is -6
3 0
3 years ago
How many dimensions exists in the universe?<br>A) 3<br>B) 5<br>C) 11<br>D) 10
nalin [4]
Your answer is 10
<span>1. Length
</span><span>2. Height
</span><span>3. Depth
</span><span>4. Time
</span><span>5. Possible Worlds
</span><span>6. A Plane of All Possible Worlds With the Same Start Conditions
</span><span>7. A Plane of All Possible Worlds With Different Start Conditions
</span><span>8. A Plane of All Possible Worlds, Each With Different Start Conditions, Each Branching Out Infinitely
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5 0
4 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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