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Sindrei [870]
3 years ago
10

Water is falling on a surface, wetting a circular area that is expanding at a rate of 6mm^2/s. How fast is the radius of the wet

ted area expanding when the radius is 109nm. (Round your answer to four decimal places.)
Mathematics
1 answer:
KengaRu [80]3 years ago
4 0

Answer:

0.0622ns

Step-by-step explanation:

Rate, R = 6mm²/s or 6 X 10∧-8m²/s

Radius, r = 109 X 10∧-9m

Area of circular object = πr² = 22/7 X (109 X 10∧-9)²= 3.734 X 10∧-14m²

Speed = 3.734 X 10∧-14m²/(6 X 10∧-8)m²/s = 0.0622ns

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Answer:

2 50-pound bags

Step-by-step explanation:

[Conversions] 50-pound bags -> 800-ounce bags

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     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

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D(c(7))
so 

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Step-by-step explanation:

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Simplify the following expression. cot^2x secx-cosx
Solnce55 [7]

ANSWER

\cos(x)   \cot ^{2} (x)

EXPLANATION

The given expression is;

\cot^{2} (x)  \sec(x)  -  \cos(x)

Change everything to

\sin(x)

and

\cos(x)

This implies that,

\frac{ \cos^{2} (x) }{ \sin^{2} (x) }  \times ( \frac{1}{ \cos(x) } )-  \cos(x)

Cancel the common factors,

\frac{ \cos(x) }{ \sin^{2} (x) }  \times ( \frac{1}{1} )-  \cos(x)

\frac{ \cos(x) }{ \sin^{2} (x) }-  \cos(x)

\frac{ \cos(x)  -  \sin ^{2} (x)  \cos(x) }{ \sin^{2} (x) }

= \frac{ \cos(x)(1  -  \sin ^{2} (x) ) }{ \sin^{2} (x) }

= \frac{ \cos(x)(\cos^{2} (x) ) }{ \sin^{2} (x) }

= \cos(x)  \cot^{2} (x)

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3 years ago
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Pepsi [2]
How long is a unit ? Is that measurement given to you ?
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