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bixtya [17]
3 years ago
9

Which expression is equivalent to (1/16)^-4

Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
4 0

Answer: =65536

Step-by-step explanation: hope this help

inysia [295]3 years ago
4 0

Answer: There are a few but: (-16)^4 is my first anwser.

Step-by-step explanation:

If you have a negative number in the parathesis you'll usually make it 1 over the number in this case 1/16 can be written as -16. That way the negative 4 can become positive four once you change the 1/16 to -16.

I hope that makes sense.

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loris [4]

(a) It looks like the ODE is

<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)

which is separable:

d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2)   =>   d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>

Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :

∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get

∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

<em>t</em> = 2<em>x</em> ^2 + <em>C</em>

arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>

<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)

(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.

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Step-by-step explanation:

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Step-by-step explanation:

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7 0
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