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lesya [120]
3 years ago
6

PLEASE HELP ILL GIVE MEDALS AND MARK BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!! NEEDS TO BE ALEGABRA 2 MEATHOD!!!!!!!!

Mathematics
1 answer:
svet-max [94.6K]3 years ago
8 0
Solve (x+3) / 3x > 2 . [I guess 3x divides (x+3) and not 3], if so, then
(x+3) > 6x → 3 > 6x-x → OR x < 3/5  OR x<0.6



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To change the shape of a function, you need to stretch or _It.
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Answer:

compress

Step-by-step explanation:

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Need help with Calculus 1 inverse trig functions
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Answer:

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

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\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}

Where u is a function of x as provided:

y=3tan^{-1}(x+\sqrt{1+x^2})

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u=(x+\sqrt{1+x^2})

Then

\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}

\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}

Taking the derivative of y

y'=3[tan^{-1}(x+\sqrt{1+x^2})]'

Using the change of variables

\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}

\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}

Operating

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}

\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}

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3 years ago
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3 to the -3 power.. ...
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The diameter of a cylinder is 4 centimeters. The height is 12 centimeters. Using 3.14 for pi, what is the volume of the cylinder
erica [24]

Answer:

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Step-by-step explanation:

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Since it is already rounded to the nearest hundredth, that is your answer.

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