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.
Answer:
"D" only 1 & 2
Step-by-step explanation:
Pythagoras theorem is a basic relationship in geometry between the three sides of a right triangle. The height from the ground to the top of the ramp 6. 0 ft. Option D is correct.
<h3>What is the Pythagoras theorem?</h3>
Pythagoras theorem is a basic relationship in geometry between the three sides of a right triangle.
It indicates that the area of the hypotenuse square is equal to the sum of the areas of the squares on the other two sides.
The given data in the problem will be;
L is the length of ramp= 10 feet
x is the horizontal distance from the bottom of the ramp = 8 feet
h is the height from the ground to the top of the ramp=?
According to Pythagoras theorem,

Hence the height from the ground to the top of the ramp 6. 0 ft. Option D is correct.
To learn more about the Pythagoras theorem refer to the link;
brainly.com/question/343682
<span>Simplifying
y(2x + 3y)(2x + 3y)
Multiply (2x + 3y) * (2x + 3y)
y(2x * (2x + 3y) + 3y * (2x + 3y))
y((2x * 2x + 3y * 2x) + 3y * (2x + 3y))
Reorder the terms:
y((6xy + 4x2) + 3y * (2x + 3y))
y((6xy + 4x2) + 3y * (2x + 3y))
y(6xy + 4x2 + (2x * 3y + 3y * 3y))
y(6xy + 4x2 + (6xy + 9y2))
Reorder the terms:
y(6xy + 6xy + 4x2 + 9y2)
Combine like terms: 6xy + 6xy = 12xy
y(12xy + 4x2 + 9y2)
(12xy * y + 4x2 * y + 9y2 * y)
(12xy2 + 4x2y + 9y3)</span>