Consider the helix below:
![r(t)= (\cos(7t), \sin(7t), -t)](https://tex.z-dn.net/?f=r%28t%29%3D%20%28%5Ccos%287t%29%2C%20%5Csin%287t%29%2C%20-t%29)
We have to determine the value of helix at t = ![\frac{\pi}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B6%7D)
So, ![r(\frac{\pi}{6})](https://tex.z-dn.net/?f=r%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29)
=![(\cos(\frac{7 \pi}{6}), \sin(\frac{7 \pi}{6}), -\frac{\pi}{6})](https://tex.z-dn.net/?f=%28%5Ccos%28%5Cfrac%7B7%20%5Cpi%7D%7B6%7D%29%2C%20%5Csin%28%5Cfrac%7B7%20%5Cpi%7D%7B6%7D%29%2C%20-%5Cfrac%7B%5Cpi%7D%7B6%7D%29)
Consider ![\cos(\frac{7 \pi}{6}) = \cos(\pi + \frac{\pi}{6}) = - \cos (\frac{\pi}{6}) = \frac{-\sqrt3}{2}](https://tex.z-dn.net/?f=%5Ccos%28%5Cfrac%7B7%20%5Cpi%7D%7B6%7D%29%20%3D%20%5Ccos%28%5Cpi%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20-%20%5Ccos%20%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B-%5Csqrt3%7D%7B2%7D)
Consider ![\sin(\frac{7 \pi}{6}) = \sin(\pi + \frac{\pi}{6}) = - \sin (\frac{\pi}{6}) = \frac{-1}{2}](https://tex.z-dn.net/?f=%5Csin%28%5Cfrac%7B7%20%5Cpi%7D%7B6%7D%29%20%3D%20%5Csin%28%5Cpi%20%2B%20%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20-%20%5Csin%20%28%5Cfrac%7B%5Cpi%7D%7B6%7D%29%20%3D%20%5Cfrac%7B-1%7D%7B2%7D)
So, the value of helix
.
Answer:
2 x² - 6x -5
Step-by-step explanation:
<u><em>Expansion:-</em></u>
Given that the expression
(8x²-7x-2)-(6x²-x+3)
= 8x² - 6x² - 7x +x -2 -3
= 2 x² - 6x -5
82,000
Hope i helped today!
-Good Luck!
Answer:
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Step-by-step explanation:
Let
S = 2b/(b+a)^2 + 2a/(b^2-a^2) factor denominator
= 2b/(b+a)^2 + 2a/((b+a)(b-a)) factor denominators
= 1/(b+a) ( 2b/(b+a) + 2a/(b-a)) find common denominator
= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a)) expand
= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a)) simplify & factor
= 2/(b+a)(b^2+a^2)/((b+a)(b-a)) simplify & rearrange
= 2(b^2+a^2)/((b+a)^2(b-a))
Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2
Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3
Answer:
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