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2 years ago

The equation y = 3.5x represents the rate, in miles

Mathematics

2 answers:

Solnce55 [7]2 years ago

7 0

Answer:

Piyush walks faster than Laura

Step-by-step explanation:

Piyush walks faster than Laura

avanturin [10]2 years ago

5 0

Answer:

Piyush walks faster than Laura

Step-by-step explanation:

we know that

The rate or speed is equal to the slope of the linear equation

<em>Laura</em>

y=3.5x

The slope is $3.5\ \frac{mi}{h}$

<em>Piyush</em>

Determine the slope of the line

we have the points

(0,0) and (4,18)

The line represent a direct variation (because passes through the origin)

$m=\frac{y}{x}$

For x=4 h, y=18 mi

substitute

$m=\frac{18}{4}=4.5\ \frac{mi}{h}$

Compare the slopes

$4.5\ \frac{mi}{h} > 3.5\ \frac{mi}{h}$

therefore

Piyush walks faster than Laura

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3 years ago

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Water is added to a cylindrical tank of radius 5 m and height of 10 m at a rate of 100 L/min. Find the rate of change of the wat

nirvana33 [79]

Answer:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

Step-by-step explanation:

For a tank similar to a cylinder the volume is given by:

$V = \pi r^2 h$

For this case we know that $r=5m$ represent the radius, $h = 10m$ the height and the rate given is:

$\frac{dV}{dt}= \frac{100 L}{min}$

For this case we want to find the rate of change of the water level when h =6m so then we can derivate the formula for the volume and we got:

$\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}$

And solving for $\frac{dh}{dt}$ we got:

$\frac{dh}{dt}= \frac{\frac{dV}{dt}}{\pi r^2}$

We need to convert the rate given into m^3/min and we got:

$Q = 100 \frac{L}{min} *\frac{1m^3}{1000L}= 0.1 \frac{m^3}{min}$

And replacing we got:

$\frac{dh}{dt}=\frac{0.1 m^3/min}{\pi (5m)^2}= 0.0012732 \frac{m}{min}$

And that represent $0.127 \frac{cm}{min}$

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3 years ago

226Ra has a half-life of 1599 years. How much is left after 1000 years if the initial amount was 10 g?

Ray Of Light [21]

$\bf \textit{Amount for Exponential Decay using Half-Life}\\\\
A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{initial amount}\to &10\\
t=\textit{elapsed time}\to &1000\\
h=\textit{half-life}\to &1599
\end{cases}
\\\\\\
A=10\left( \frac{1}{2} \right)^{\frac{1000}{1599}}$

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2 years ago