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erastovalidia [21]
3 years ago
9

Find the distance between each pair of points. Round to the nearest tenth, if necessary.

Mathematics
1 answer:
Westkost [7]3 years ago
3 0
\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
O&({{ 0}}\quad ,&{{ 0}})\quad 
%  (c,d)
P&({{ -5}}\quad ,&{{ 6}})\\
K&({{ 5}}\quad ,&{{ 0}})\quad 
%  (c,d)
L&({{ -2}}\quad ,&{{ 1}})

\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
OP=\sqrt{(-5-0)^2+(6-0)^2}\qquad \qquad KL=\sqrt{(-2-5)^2+(1-0)^2}
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6 0
2 years ago
Please help with math problem give 5 star if do
Juliette [100K]

Answer:

C : -6

Step-by-step explanation:

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(C)

Step-by-step explanation:

The volume of the conical pile is given by

V = \dfrac{\pi}{3}r^2h

Taking the derivative of V with respect to time, we get

\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)

\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)

Since r is always equal to h, we can set

\dfrac{dr}{dt} = \dfrac{dh}{dt}

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\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)

\:\:\:\:\:\:\:= \pi r^2\dfrac{dh}{dt}

Solving for dh/dt, we get

\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}

\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})

\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}

7 0
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VikaD [51]

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