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choli [55]
3 years ago
13

The area of a circular wave expands across a still pond such that its radius increases by 14 cm each second. Write a formula for

the area A of the circle as a function of time t since the wave begins: A=
Mathematics
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

A = π(14t)²

Step-by-step explanation:

The radius is increasing at the rate of 14 cm per second.

We need to find the formula for the area A of the circle as the function of time t.

Initial area of the circle,

A = πr², where r is the radius of the circle

Area as a function of t will be :

A = π(14t)²

Here, 14t is the radius of the wave.

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Reflecting diagonally gives you the grayed-out shape, and translating it 3 units down will give you Figure E.

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14. Anna, Bella, Claire, Dora and Erika meet at a party. Each pair who know each other shake hands exactly once.
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The equation y=3x2-3 is graphed below on Graph A what is the equation graphed on graph B?
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4 0
4 years ago
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Guys need help its surd one <br><img src="https://tex.z-dn.net/?f=5%20%5Csqrt%7B3%20%5Cdiv%202%20%5Csqrt%7B3%20-%20%20%5Csqrt%7B
sergey [27]

Answer:

Step-by-step explanation:

\displaystyle \Large \boldsymbol{} 5\cdot  \sqrt{3:2\sqrt{3-\sqrt{2} } }  =\\\\\\ 5\cdot  \sqrt{\frac{3}{2\sqrt{3-\sqrt{2} } } \cdot \frac{2\sqrt{3+\sqrt{2} } }{2\sqrt{3+\sqrt{2} } } } = \\\\\\5\cdot  \sqrt{\frac{6\sqrt{3+\sqrt{2} } }{4\cdot \sqrt{9-2} } }  =5\sqrt{\frac{6\sqrt{3+\sqrt{2} } }{2\sqrt{7} } }

But if the condition would be like this

\displaystyle \Large \boldsymbol{} 5\cdot  \sqrt{3:2\sqrt{3-\boldsymbol2\sqrt{2} } }  =\\\\\\5\cdot \sqrt{\frac{3}{2\cdot \sqrt{\underbrace{2+1-2\sqrt{2} \cdot \sqrt{1} }_{(\sqrt{2} -1)^2} } }} =\\\\\\5\cdot \sqrt{\frac{3}{2\cdot (\sqrt{2}-1) } } } = \\\\\\ 5\cdot  \sqrt{\frac{3}{2(\sqrt{2}-1 )}  \cdot \frac{\sqrt{2} +1}{\sqrt{2}+1 }}   =5 \sqrt{\frac{3}{2(2-1)} }  =\boxed{5\sqrt{1,5} }

8 0
3 years ago
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