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Evgesh-ka [11]
3 years ago
15

What is the sum of prime numbers between 40 and 56​

Mathematics
2 answers:
Naddika [18.5K]3 years ago
4 0

Answer:

184

Step-by-step explanation:

The prime numbers between 40 and 56​ are 41, 43, 47, and 53.

Add the prime numbers.

41 + 43 + 47 + 53

= 184

levacccp [35]3 years ago
3 0

Answer:

184

Step-by-step explanation:

Prime numbers between 40 and 56: 41, 43, 47, 53.

41+43+47+53=

84+47+53=

131+53=

184

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One of the same side angles of two parallel lines is 20° smaller than the other one. Find all angles.
Elden [556K]

Answer: 80 and 100


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Prove Euler's identity using Euler's formula.<br> e^ix = cos x + i sin x
Korolek [52]

First list all the terms out.

e^ix = 1 + ix/1! + (ix)^2/2! + (ix)^3/3! ...

Then, we can expand them.

e^ix = 1 + ix/1! + i^2x^2/2! + i^3x^3/3!...

Then, we can use the rules of raising i to a power.

e^ix = 1 + ix - x^2/2! - ix^3/3!...

Then, we can sort all the real and imaginary terms.

e^ix = (1 - x^2/2!...) + i(x - x^3/3!...)

We can simplify this.

e^ix = cos x + i sin x

This is Euler's Formula.

What happens if we put in pi?

x = pi

e^i*pi = cos(pi) + i sin(pi)

cos(pi) = -1

i sin(pi) = 0

e^i*pi = -1 OR e^i*pi + 1 = 0

That is Euler's identity.

3 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!

6 0
3 years ago
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