At start ( t = 0 nanoseconds ) :
E ( t = 0 ) = 2.645 J
E ( t = 1 ) = 6.290 J
E ( t = 1 ) : E ( t = 0 ) = 6.290 : 2.645 = 2.37
Also:
E ( t = 2 ) : E ( t = 1 ) = 14.909 : 6.290 = 2.37
E ( t = 3 ) ; E ( t = 2 ) = 35.335 : 14.909 = 2.37
Therefore, the formula for calculating the energy of the system is:
E ( t ) = 2.645 * 2.37 ^ t
The answer is 4) exponential growth.
The product of 5 and 2/3 is 5.6666
For the equation F(x) = ax² + bx + c we have:
- maximum value if a<0
- minimum value if a>0
F(x) = -3x² + 18x + 3 ⇒ a = -3, b = 18
a < 0 ⇒ the function has a maximum value
Quadratic function has the maximum value (or minimum) at vertex of its parabola.
The maximum value is k at x=h where:
and k = F(h)
![h=\dfrac{-18}{2\cdot(-3)}=3\\\\F(3)=-3\cdot3^2+18\cdot3+3=-27+54+3=30](https://tex.z-dn.net/?f=h%3D%5Cdfrac%7B-18%7D%7B2%5Ccdot%28-3%29%7D%3D3%5C%5C%5C%5CF%283%29%3D-3%5Ccdot3%5E2%2B18%5Ccdot3%2B3%3D-27%2B54%2B3%3D30)
Therefore:
<h3>
The function has a maximum value of 30 at x = 3</h3>
Answer:
16![\pi](https://tex.z-dn.net/?f=%5Cpi)
Step-by-step explanation:
area of a circle is pi x r x r
since it is a quarter
1/4 x pi x 8 x 8 = 16 x pi
Step-by-step explanation:
You said the other one would be the last haha just kidding I'm glad to help.
16. ![\frac{k}{2}=(-5)^2](https://tex.z-dn.net/?f=%5Cfrac%7Bk%7D%7B2%7D%3D%28-5%29%5E2)
First, get rid of that parenthesis.
![\frac{k}{2}=25](https://tex.z-dn.net/?f=%5Cfrac%7Bk%7D%7B2%7D%3D25)
Now multiply both sides by 2 so that you can isolate k
![2(\frac{k}{2})=25*2](https://tex.z-dn.net/?f=2%28%5Cfrac%7Bk%7D%7B2%7D%29%3D25%2A2)
![k=50](https://tex.z-dn.net/?f=k%3D50)
19. ![\frac{r}{3}=\frac{121}{11}](https://tex.z-dn.net/?f=%5Cfrac%7Br%7D%7B3%7D%3D%5Cfrac%7B121%7D%7B11%7D)
This is a pretty easy one. If you didn't know, 121/11 is actually 11 :)
![\frac{r}{3}=11](https://tex.z-dn.net/?f=%5Cfrac%7Br%7D%7B3%7D%3D11)
Simply multiply by 3 to isolate r :)
![3(\frac{r}{3})=11*3](https://tex.z-dn.net/?f=3%28%5Cfrac%7Br%7D%7B3%7D%29%3D11%2A3)
![r=33](https://tex.z-dn.net/?f=r%3D33)