The answer is 0.1 ❤️❤️❤️❤️ hope this helps
Answer:
One mole of lithium (Li) is equal to 6.022 X 1023 atoms of lithium.
Answer:
Problem 20)
![\displaystyle \frac{dy}{dx}=(\cos x)^x\left(\ln \cos x-x\tan x\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D%28%5Ccos%20x%29%5Ex%5Cleft%28%5Cln%20%5Ccos%20x-x%5Ctan%20x%5Cright%29)
Problem 21)
A)
The velocity function is:
![\displaystyle v(t) =2\pi(\cos(2\pi t)-\sin(\pi t))](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20v%28t%29%20%3D2%5Cpi%28%5Ccos%282%5Cpi%20t%29-%5Csin%28%5Cpi%20t%29%29)
The acceleration function is:
![\displaystyle a(t)=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%28t%29%3D-2%5Cpi%5E2%282%5Csin%282%5Cpi%20t%29%2B%5Ccos%28%5Cpi%20t%29%29)
B)
![s(0)=2\text{, }v(0) = 2\pi \text{ m/s}\text{, and } a(0) = -2\pi^2\text{ m/s$^2$}](https://tex.z-dn.net/?f=s%280%29%3D2%5Ctext%7B%2C%20%7Dv%280%29%20%3D%202%5Cpi%20%5Ctext%7B%20m%2Fs%7D%5Ctext%7B%2C%20and%20%7D%20a%280%29%20%3D%20-2%5Cpi%5E2%5Ctext%7B%20m%2Fs%24%5E2%24%7D)
Step-by-step explanation:
Problem 20)
We want to differentiate the equation:
![\displaystyle y=\left(\cos x\right)^x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D%5Cleft%28%5Ccos%20x%5Cright%29%5Ex)
We can take the natural log of both sides. This yields:
![\displaystyle \ln y = \ln((\cos x)^x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cln%20y%20%3D%20%5Cln%28%28%5Ccos%20x%29%5Ex%29)
Since ln(aᵇ) = bln(a):
![\displaystyle \ln y =x\ln \cos x](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cln%20y%20%3Dx%5Cln%20%5Ccos%20x)
Take the derivative of both sides with respect to <em>x: </em>
<em />
<em />
Implicitly differentiate the left and use the product rule on the right. Therefore:
![\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \cos x+x\left(\frac{1}{\cos x}\cdot -\sin(x)\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7By%7D%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cln%20%5Ccos%20x%2Bx%5Cleft%28%5Cfrac%7B1%7D%7B%5Ccos%20x%7D%5Ccdot%20-%5Csin%28x%29%5Cright%29)
Simplify:
![\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \cos x-\frac{x\sin x}{\cos x}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B1%7D%7By%7D%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Cln%20%5Ccos%20x-%5Cfrac%7Bx%5Csin%20x%7D%7B%5Ccos%20x%7D)
Simplify and multiply both sides by <em>y: </em>
<em />
<em />
Since <em>y</em> = (cos x)ˣ:
Problem 21)
We are given the position function of a particle:
![\displaystyle s(t)= \sin (2\pi t)+2\cos(\pi t)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20s%28t%29%3D%20%5Csin%20%282%5Cpi%20t%29%2B2%5Ccos%28%5Cpi%20t%29)
A)
Recall that the velocity function is the derivative of the position function. Hence:
![\displaystyle v(t)=s'(t)=\frac{d}{dt}[\sin(2\pi t)+2\cos(\pi t)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%28t%29%3Ds%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%282%5Cpi%20t%29%2B2%5Ccos%28%5Cpi%20t%29%5D)
Differentiate:
![\displaystyle \begin{aligned} v(t) &= 2\pi \cos(2\pi t)-2\pi \sin(\pi t)\\&=2\pi(\cos(2\pi t)-\sin(\pi t))\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20v%28t%29%20%26%3D%202%5Cpi%20%5Ccos%282%5Cpi%20t%29-2%5Cpi%20%5Csin%28%5Cpi%20t%29%5C%5C%26%3D2%5Cpi%28%5Ccos%282%5Cpi%20t%29-%5Csin%28%5Cpi%20t%29%29%5Cend%7Baligned%7D)
The acceleration function is the derivative of the velocity function. Hence:
![\displaystyle a(t)=v'(t)=\frac{d}{dt}[2\pi(\cos(2\pi t)-\sin(\pi t))]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%28t%29%3Dv%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B2%5Cpi%28%5Ccos%282%5Cpi%20t%29-%5Csin%28%5Cpi%20t%29%29%5D)
Differentiate:
![\displaystyle \begin{aligned} a(t)&=2\pi[-2\pi\sin(2\pi t)-\pi\cos(\pi t)]\\&=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20a%28t%29%26%3D2%5Cpi%5B-2%5Cpi%5Csin%282%5Cpi%20t%29-%5Cpi%5Ccos%28%5Cpi%20t%29%5D%5C%5C%26%3D-2%5Cpi%5E2%282%5Csin%282%5Cpi%20t%29%2B%5Ccos%28%5Cpi%20t%29%29%5Cend%7Baligned%7D)
B)
The position at <em>t</em> = 0 will be:
![\displaystyle \begin{aligned} s(0)&=\sin(2\pi(0))+2\cos(\pi(0))\\&=\sin(0)+2\cos(0)\\&=(1)+2(1)\\&=2\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20s%280%29%26%3D%5Csin%282%5Cpi%280%29%29%2B2%5Ccos%28%5Cpi%280%29%29%5C%5C%26%3D%5Csin%280%29%2B2%5Ccos%280%29%5C%5C%26%3D%281%29%2B2%281%29%5C%5C%26%3D2%5Cend%7Baligned%7D)
The velocity at <em>t</em> = 0 will be:
![\displaystyle \begin{aligned} v(0)&=2\pi(\cos(2\pi (0)-\sin(\pi(0))\\&=2\pi(\cos(0)-\sin(0))\\&=2\pi((1)-(0))\\&=2\pi \text{ m/s}\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20v%280%29%26%3D2%5Cpi%28%5Ccos%282%5Cpi%20%280%29-%5Csin%28%5Cpi%280%29%29%5C%5C%26%3D2%5Cpi%28%5Ccos%280%29-%5Csin%280%29%29%5C%5C%26%3D2%5Cpi%28%281%29-%280%29%29%5C%5C%26%3D2%5Cpi%20%5Ctext%7B%20m%2Fs%7D%5Cend%7Baligned%7D)
And the acceleration at <em>t</em> = 0 will be:
![\displaystyle \begin{aligned} a(0) &= -2\pi ^2(2\sin(2\pi(0))+\cos(\pi(0)) \\ & = -2\pi ^2(2\sin(0)+\cos(0)) \\ &= -2\pi ^2(2(0)+(1)) \\ &= -2\pi^2(1) \\ &= -2\pi^2\text{ m/s$^2$} \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20a%280%29%20%26%3D%20-2%5Cpi%20%5E2%282%5Csin%282%5Cpi%280%29%29%2B%5Ccos%28%5Cpi%280%29%29%20%5C%5C%20%26%20%3D%20-2%5Cpi%20%5E2%282%5Csin%280%29%2B%5Ccos%280%29%29%20%5C%5C%20%26%3D%20-2%5Cpi%20%5E2%282%280%29%2B%281%29%29%20%5C%5C%20%26%3D%20-2%5Cpi%5E2%281%29%20%5C%5C%20%26%3D%20-2%5Cpi%5E2%5Ctext%7B%20m%2Fs%24%5E2%24%7D%20%5Cend%7Baligned%7D)
Answer:
C. (13,8)
Step-by-step explanation:
<u>Midpoint formula is (x²+x¹</u><u> </u><u>/2 , y²+y¹</u><u> </u><u>/2 )</u>
Midpoint= L(3,5)
J(-7 , 2) ; K(x , y)
(3,5) = (x+-7 /2 , y+2 /2)
x-7 /2 = 3
x-7 = 6
x= 13
y+2 /2 =5
y+2 = 10
y= 8
Therefore K is K(13, 8)