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nadezda [96]
3 years ago
10

3)

Mathematics
1 answer:
vova2212 [387]3 years ago
8 0

Answer:

C. Negative

Step-by-step explanation:

The line is not horizontal or vertical, therefore we the slope cannot be zero or undefined.

This leaves us with either positive or negative...

A positive slope goes from the lower left corner (Quadrant 3) of the graph up to the upper right (Quadrant 1).

A negative slope goes from the upper left corner (Quadrant 2) of the graph down to the lower right (Quadrant 4).

Since this line goes from the upper left corner to the lower right, it has a negative slope

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10. simplify the rational expression by rationalizing the denominator.( 1 point ) 4√150/√189x
ddd [48]
Rationalizing the denominator, simply means "getting rid of that pesky root at the bottom", and we do so by simply multiplying it by something to take it out, of course, we multiply the bottom, we have to also multiply the top,

\bf \cfrac{4\sqrt{150}}{\sqrt{189x}}\cdot \cfrac{\sqrt{189x}}{\sqrt{189x}}\implies \cfrac{4\sqrt{150}\sqrt{189x}}{(\sqrt{189x})^2}\implies \cfrac{4\sqrt{(150)({189x})}}{189x}

\bf \cfrac{4\sqrt{28350x}}{189x}\qquad 
\begin{cases}
28350=2\cdot 3\cdot 3\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7\\
\qquad 2\cdot 3^2\cdot 3^2\cdot 5^2\cdot 7\\
\qquad 2\cdot (3^2)^2\cdot 5^2\cdot 7\\
\qquad 2\cdot (3^2\cdot 5)^2\cdot 7\\
\qquad 2\cdot (45)^2\cdot 7\\
\qquad 14\cdot 45^2
\end{cases}\implies \cfrac{4\sqrt{14\cdot 45^2}}{189x}
\\\\\\
\cfrac{4\cdot 45\sqrt{14}}{189}\implies \cfrac{180\sqrt{14}}{189x}\implies \cfrac{20\sqrt{14}}{21x}
4 0
3 years ago
Solve the system of equations. <br> 9x+11y=2 <br> 3x+4y=4
Ronch [10]

Answer:

X= -12 Y=10

Step-by-step explanation:

4 0
3 years ago
Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^
Gemiola [76]

Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

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3 years ago
I NEED HELP WITH A PROBLEM ​
postnew [5]

Answer:

the first one I guess

Step-by-step explanation:

6 0
3 years ago
Students roll a 6-sided number cube and spin the shown spinner. blue blue orange red orange red orange purple What is the probab
Ede4ka [16]

Answer:

1. 8.333% 2. I don't know ( sorry )

6 0
3 years ago
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