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Sidana [21]
3 years ago
6

10 points!! Will mark brainliest!!! Please help me I don’t understand this very well

Mathematics
2 answers:
Lisa [10]3 years ago
5 0
Hope that kinda helps.

Katen [24]3 years ago
4 0

Answer:

9?

Step-by-step explanation:

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i forgot how to find equivalent fractions. i know that the bottom would be times by 3 to get 18 but if you times 3 by 3 it would
tester [92]
Set up a proportion.  take 3/6 = 20/x
Cross multiply 3 * x = 6 * 20
Simplify  3x = 120
Divide by 3  x = 40
So the denominator of the bottom fraction is 40
6 0
3 years ago
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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!

6 0
3 years ago
What is the equation of a line going through points (-15,70) and (5,10)
andrey2020 [161]
Simple....

you have:

(-15,70) and (5,10)

Using-->>

\frac{  y_{1} -y_{2} }{ x_{1}-  x_{2} }

\frac{70-10}{-15-5} = \frac{60}{-20} = -3

Using y=mx+b...and any of the coordinates....

70= -3(-15)+b

70=45+b

70=45+b
-45  -45

25=b

y=-3x+25

Thus,your answer.

8 0
3 years ago
4 teammates share 5 granola bars equally. How much of a bar dows each person get?
worty [1.4K]
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6 0
3 years ago
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Romashka-Z-Leto [24]

Answer:

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Step-by-step explanation:

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