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3 years ago

11

Find the critical numbers sin^2x+cosx 0 < x< 2pi

1 answer:

ryzh [129]3 years ago

5 0

$\bf sin^2(x)+cos(x)\iff [sin(x)]^2+cos(x)\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=2sin(x)cos(x)-sin(x)\implies 0=2sin(x)cos(x)-sin(x)
\\\\\\
0=sin(x)[2cos(x)-1]\implies 
\begin{cases}
0=sin(x)\\
sin^{-1}(0)=\measuredangle x\\
\pi \\
----------\\
0=2cos(x)-1\\\\ \frac{1}{2}=cos(x)\\\\ cos^{-1}\left( \frac{1}{2} \right)=\measuredangle x\\\\ \frac{\pi }{3},\frac{5\pi }{3}
\end{cases}$

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For what value of x is sq 896z^15/225z^6 = xz^4/15

Ratling [72]

Answer:

Value of x is, 8

Step-by-step explanation:

Given: $\sqrt{\frac{896z^{15}}{225z^6}} =\frac{xz^4}{15} \sqrt{14z}$ ,.....[1]

To find the value of x:

Using exponent rules:

$(a^n)^m = a^{nm}$

$a^n \cdot a^m = a^{n+m}$

Taking square both sides in [1] we have;

$\frac{896z^{15}}{225z^6} = (\frac{xz^4}{15})^2 \cdot (14z)$

Simplify:

$\frac{896z^{15}}{225z^6} =\frac{x^2z^8}{225} \cdot (14z)$

or

$\frac{896z^{15}}{225z^6} =\frac{14x^2z^9}{225}$

Multiply both sides by $\frac{225}{14z^9}$ we get;

$x^2 = \frac{896 z^{15}}{225z^6} \times \frac{225}{14 z^9} = \frac{896 z^{15}}{14 z^{9+6}}$

Simplify:

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Simplify:

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alisha [4.7K]

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Step-by-step explanation:

The maximum concentration will happen in t hours. t is found when

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Applying the quotient derivative formula

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