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2 years ago

Help,anyone can help me do quetion.

Mathematics

2 answers:

Cerrena [4.2K]2 years ago

6 0

Answer:

Step-by-step explanation:

angle y +60 = 180 degree (being adjacent angles of a trapezium )

angle y =180 - 60

angle y =120 degree

angle x+30 degree = 180 degree (being opposite angles of a trapezium)

angle x =180 - 30

angle x=150 degree

75 + 145 + 90 + y = 360 degree (sum of angles of a quadilateral is 360 degree)

310 + y = 360

y = 360 - 310

y = 50 degree

mel-nik [20]2 years ago

3 0

X=60 and Y=30
A quadrilateral with 2 parallel lines is a parallelogram which has equal angle to the opposite angle

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$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.