Answer:
Step-by-step explanation:
We have:
And we want to find B’(6).
So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:
![\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B24.6%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D)
We can move the constant outside:
![\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%28t%29%3D24.6%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D)
Now, we will utilize the product rule. The product rule is:
We will let:
Then:
(The derivative of u was determined using the chain rule.)
Then it follows that:
![\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20B%5E%5Cprime%28t%29%26%3D24.6%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D%20%5C%5C%20%5C%5C%20%26%3D24.6%5B%28%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%29%288-t%29%20-%20%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%5D%20%5Cend%7Baligned%7D)
Therefore:
![\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%286%29%20%3D24.6%5B%28%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B%5Cpi%20%286%29%7D%7B10%7D%29%29%288-%286%29%29-%20%5Csin%28%5Cfrac%7B%5Cpi%20%286%29%7D%7B10%7D%29%5D)
By simplification:
![\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%286%29%3D24.6%20%5B%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B3%5Cpi%7D%7B5%7D%29%282%29-%5Csin%28%5Cfrac%7B3%5Cpi%7D%7B5%7D%29%5D%20%5Capprox%20-28.17)
So, the slope of the tangent line to the point (6, B(6)) is -28.17.
I’m not 100% sure but I think it’s 4 for the gcf
Answer:
Solutions are 2, -1 + 0.5 sqrt10 i and -1 - 0.5 sqrt10 i
or 2, -1 + 1.58 i and -1 - 1.58i
(where the last 2 are equal to nearest hundredth).
Step-by-step explanation:
The real solution is x = 2:-
x^3 - 8 = 0
x^3 = 8
x = cube root of 8 = 2
Note that a cubic equation must have a total of 3 roots ( real and complex in this case). We can find the 2 complex roots by using the following identity:-
a^3 - b^3 = (a - b)(a^2 + ab + b^2).
Here a = x and b = 2 so we have
(x - 2)(x^2 + 2x + 4) = 0
To find the complex roots we solve x^2 + 2x + 4 = 0:-
Using the quadratic formula x = [-2 +/- sqrt(2^2 - 4*1*4)] / 2
= -1 +/- (sqrt( -10)) / 2
= -1 + 0.5 sqrt10 i and -1 - 0.5 sqrt10 i
Answer:
b
Step-by-step explanation:
