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finlep [7]
3 years ago
14

Pre cacl pleaseeeee help. the options are in the attachments as well

Mathematics
1 answer:
ivanzaharov [21]3 years ago
3 0

Answer:

The values of x so that y = \csc x have vertical asymptotes are -2\pi, -\pi, 0, \pi, 2\pi.

Step-by-step explanation:

The function cosecant is the reciprocal of the function sine and vertical asymptotes are located at values of x so that function cosecant becomes undefined, that is, when function sine is zero, whose periodicity is \pi. Then, the  vertical asymptotes associated with function cosecant are located in the values of x of the form:

x = 0\pm \pi\cdot i, \forall \,i\in \mathbb{N}_{O}

In other words, the values of x so that y = \csc x have vertical asymptotes are -2\pi, -\pi, 0, \pi, 2\pi.

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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
Solve:<br>p+ 7- 5= -10 - 3<br>please use an explanation so I know how to do this next time pleaasee​
skelet666 [1.2K]

Answer:

P = - 15

Step-by-step explanation:

P+7-5 = -10 - 3

P +2 = - 13

(If taking from one side,you must do to the other to get P alone.)

P+2 = - 13

-2 = - 2

The positive 2, minus the negative 2 cancels each other out, now we must subtract 2 from our other side. - 13 - 2 = - 15

P = -15

7 0
3 years ago
A ruler is used to measure the width of the spine of a book.
sp2606 [1]

Answer:

1.00 cm Is the greatest level of precision possible for the measurement of the width of the spin of the book

3 0
3 years ago
What is the greatest common factor of 48 and 168
snow_lady [41]

Answer:

24

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Solve x+2.4=43 do the inverse to find the answer
yulyashka [42]
X+2.4=43
-2.4 - 2.4
———————
X= 40.6
6 0
3 years ago
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