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3 years ago

Can someone help me with this question

Mathematics

1 answer:

horsena [70]3 years ago

6 0

Step-by-step explanation:

Given: diameter = 16 in

hence radius, r is 16 ÷ 2 = 8 in

a) notice that the fan blades divide the circle into 3 equal sectors.

Each sector is hence 1/3 of the entire circumference of the circle

Arc Length of each sector (between blades),

= (1/3) x circumference

= (1/3) x 2πr

= (1/3) x 2 x 3.142 x 8

= 16.76 in (Answer)

b) similarly, the area between the blades

= (1/3) x πr²

= (1/3) x 3.142 x 8²

= 67.03 in² (answer)

c) The angular velocity is the radians per second that the tip moves

It is given that it moves 8π radians in 2 seconds

hence in 1 second, we expect the tip to move,

8π ÷ 2 = 4π rad/s (Answer)

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2 years ago

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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 $y = e^x$

-----------------------

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.

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2 years ago