Answer:
C) 7
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Work Shown:
Use the slope formula
m = (y2-y1)/(x2-x1)
Plug in the given slope we want m = -5/3 and also the coordinates of the points. Then isolate r
m = (y2-y1)/(x2-x1)
-5/3 = (2-r)/(r-4)
-5(r-4) = 3(2-r) .... cross multiplying
-5r+20 = 6-3r
-5r+20+5r = 6-3r+5r .... adding 5 to both sides
20 = 6+2r
20-6 = 6+2r-6 ....subtracting 6 from both sides
14 = 2r
2r = 14
2r/2 = 14/2 .... dividing both sides by 2
r = 7
The slope of the line through (4,7) and (7,2) should be -5/3, let's check that
m = (y2-y1)/(x2-x1)
m = (2-7)/(7-4)
m = -5/3
The answer is confirmed
All real numbers is the correct answer.
Answer:
Hope this helps
Step-by-step explanation:
Crossproduct is explained. You can use it to get all the values as well as being able to convert into other currencies.
Answer:
x = 35.4
Step-by-step explanation:
Sum of all interior angles in a triangle = 180°
Therefore:
(x + 2)° + (2x + 2)° + (2x - 1)° = 180°
x + 2 + 2x + 2 + 2x - 1 = 180
Collect like terms
x + 2x + 2x + 2 + 2 - 1 = 180
5x + 3 = 180
Subtract 3 from each side
5x + 3 - 3 = 180 - 3
5x = 177
Divide both sides by 5
5x/5 = 177/5
x = 35.4
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.
