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Aleks04 [339]
3 years ago
7

What is nine and forty-two hundredths

Mathematics
1 answer:
stepan [7]3 years ago
5 0

Answer:

9.42

Step-by-step explanation:

Breaking the phrase down:

'Nine' would be the number 9 in the ones place.

'And' represents the decimal in a number. ('.')

'Forty-Two Hundredths" is 0.42.

So, "nine and forty-two hundredths" would be 9.42.

Hope this helps.

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Zigmanuir [339]

Answer:

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Step-by-step explanation:

Angles in a triangle add up to 180 degrees.

The triangle with 39 and 90 has a missing angle of 51 degrees, and the triangle with 51 and 90 vice versa.

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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.

The derivative of the inverse tangent is given as

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

Where u is a function of x as provided:

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If we set

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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The answer is 30, hope this helps. 
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