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Nata [24]
1 year ago
6

Which expression is equivalent to RootIndex 4 StartRoot StartFraction 16 x Superscript 11 Baseline y Superscript 8 Baseline Over

81 x Superscript 7 Baseline y Superscript 6 Baseline EndFraction EndRoot? Assume x Greater-than 0 and y not-equals 0. StartFraction 4 x (RootIndex 4 StartRoot y squared EndRoot) Over 9 EndFraction StartFraction 2 x (RootIndex 4 StartRoot y squared EndRoot) Over 3 EndFraction StartFraction 4 x squared y Over 9 EndFraction StartFraction 2 x squared y Over 3 EndFraction
Mathematics
2 answers:
trasher [3.6K]1 year ago
5 0

Answer:

I said D

Step-by-step explanation:

storchak [24]1 year ago
5 0

Answer:

its d i just took the test

Step-by-step explanation:

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Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. x2 + y2 = 25 (a) Fin
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Answer:

(a) \frac{dy}{dt}=-3\frac{3}{4}

(b) \frac{dx}{dt}=3\frac{3}{4}

Step-by-step explanation:

x^{2} +y^{2}=25

Take \frac{d}{dt} of of each term.

\frac{d}{dt}(x^{2})+\frac{d}{dt}(y^{2})=\frac{d}{dt}(25)\\\\(\frac{d}{dx}(x^{2})*\frac{dx}{dt}) +(\frac{d}{dy}(y^{2})*\frac{dy}{dt})=\frac{d}{dt}(25)\\\\2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0\\\\

For Question a

2y\frac{dy}{dt}=-2x\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{-2x\frac{dx}{dt}}{2y} \\\\\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

Given that x = 3, y = 4, and dx/dt = 5.

\frac{dy}{dt}=-\frac{3}{4}*5=-\frac{15}{4}\\   \\\frac{dy}{dt}=-3\frac{3}{4}

For Question b

2x\frac{dx}{dt}=-2y\frac{dy}{dt}\\\\\frac{dx}{dt}=\frac{-2y\frac{dy}{dt}}{2x} \\\\\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}

Given that x = 4, y = 3, and dx/dt = -5.

\frac{dx}{dt}=-\frac{3}{4}*-5=\frac{15}{4}\\   \\\frac{dx}{dt}=3\frac{3}{4}

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2 years ago
An element with mass 730 grams decays by 27.6% per minute. How much of the element is remaining after 12 minutes, to the nearest
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\bf \qquad \textit{Amount for Exponential Decay}\\\\
A=I(1 - r)^t\qquad 
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A=\textit{accumulated amount}\\
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r=rate\to 27.6\%\to \frac{27.6}{100}\to &0.276\\
t=\textit{elapsed time}\to &12\\
\end{cases}
\\\\\\
A=730(1-0.276)^{12}\implies A=730(0.724)^{12}
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