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Monica [59]
3 years ago
7

Please help me with this

Mathematics
1 answer:
mash [69]3 years ago
4 0

Answer:

The correct answer is first option

24

Step-by-step explanation:

From the figure we get, mAXM = 72° and  m<AMR = 38°

Also it is given that, all triangles are isosceles triangles and

m<FXA = 96°

<u>To find the measure of <FXM</u>

From the figure we get,

m<FXA =  m<AXM + m<FXM

m<FXM = m<FXA - m<AXM

 = 96 - 72

 = 24

Therefore the  correct answer is first option

24

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Use the quadratic formula to find the solutions to the quadratic equation below x^2-6x-5=0
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Answer:

\large\boxed{x=3-\sqrt{14}\ \vee\ x=3+\sqrt{14}}

Step-by-step explanation:

x^2-6x-5=0\qquad\text{add 5 to both sides}\\\\x^2-6x=5\\\\x^2-2(x)(3)=5\qquad\text{add}\ 3^2\ \text{to both sides}\\\\x^2-2(x)(3)+3^2=5+3^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-3)^2=5+9\\\\(x-3)^2=14\to x-3=\pm\sqrt{14}\qquad\text{add 3 to both sides}\\\\x=3-\sqrt{14}\ \vee\ x=3+\sqrt{14}

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Are the expressions below equivalent 3(4x+8) 12x+24
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lidiya [134]

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:

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}}}

8 0
3 years ago
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