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Misha Larkins [42]
3 years ago
8

Help please!! i don’t understand this one!

Mathematics
1 answer:
12345 [234]3 years ago
5 0

Answer:

  C, D

Step-by-step explanation:

This equation can be solved any of several ways. One that doesn't require much thought is using the quadratic formula.

For ax² +bx +c = 0, the solutions are ...

  x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

In the given equation, a=2, b=11, c=5, so this becomes ...

  x=\dfrac{-11\pm\sqrt{11^2-4\cdot2\cdot5}}{2\cdot2}=\dfrac{-11\pm\sqrt{81}}{4}\\\\x=\dfrac{-11\pm9}{4}=\left\{-\dfrac{20}{4},-\dfrac{2}{4}\right\}=\{-5,-\frac{1}{2}\}

The solutions are ...

  C. x = -5

  D. x = -1/2

_____

My personal favorite is using a graphing calculator. The solutions are the x-intercepts of the expression on the left. That is, where its value is zero, as the equation says.

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\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:

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

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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