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3 years ago

A circular fishpond has a result of 2.5m what is the area of the fishpond?

Mathematics

1 answer:

sattari [20]3 years ago

8 0

Answer:

$19.63 {m}^{2}$

<h3>3rd answer is correct</h3>

Step-by-step explanation:

$\pi {r}^{2} \\ \pi \times 2.5 \times 2.5 \\ = 19.63 {m}^{2}$

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3 years ago

An airliner maintaining a constant elevation of 2 miles passes over an airport at noon traveling 500 mi/hr due west. At 1:00 PM,

butalik [34]

Answer:

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

Step-by-step explanation:

Let suppose that airliners travel at constant speed. The equations for travelled distance of each airplane with respect to origin are respectively:

First airplane

$r_{A} = 500\,\frac{mi}{h}\cdot t\\r_{B} = 550\,\frac{mi}{h}\cdot t$

Where t is the time measured in hours.

Since north and west are perpendicular to each other, the staight distance between airliners can modelled by means of the Pythagorean Theorem:

$s=\sqrt{r_{A}^{2}+r_{B}^{2}}$

Rate of change of such distance can be found by the deriving the expression in terms of time:

$\frac{ds}{dt}=\frac{r_{A}\cdot \frac{dr_{A}}{dt}+r_{B}\cdot \frac{dr_{B}}{dt}}{\sqrt{r_{A}^{2}+r_{B}^{2}} }$

Where $\frac{dr_{A}}{dt} = 500\,\frac{mi}{h}$ and $\frac{dr_{B}}{dt} = 550\,\frac{mi}{h}$ , respectively. Distances of each airliner at 2:30 PM are:

$r_{A}= (500\,\frac{mi}{h})\cdot (1.5\,h)\\r_{A} = 750\,mi$

$r_{B}=(550\,\frac{mi}{h} )\cdot (1.5\,h)\\r_{B} = 825\,mi$

The rate of change is:

$\frac{ds}{dt}=\frac{(750\,mi)\cdot (500\,\frac{mi}{h} )+(825\,mi)\cdot(550\,\frac{mi}{h})}{\sqrt{(750\,mi)^{2}+(825\,mi)^{2}} }$

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

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3 years ago

X^2+10x-6=0.........

Georgia [21]

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3 years ago

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My name is Ann [436]

Answer:

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Step-by-step explanation:

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3 years ago

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miss Akunina [59]

Answer:

678.74

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3 years ago