J, K, and L are collinear
J is between K and L
LK = KJ + LJ
LK = 9x + 7
KJ = 2x - 3
LJ = 4x - 8
9x + 7 = (2x - 3) + (4x - 8)
9x + 7 = (2x + 4x) + (-3 - 8)
9x + 7 = 6x + -11
9x + 7 = 6x - 11
9x + 7 - 7 = 6x - 11 - 7
9x = 6x - 18
9x - 6x = 6x - 6x - 18
3x = -18
x = -6
∴ The value of x is -6
Well, to find u, we have to remove all that is attached to it so the equation can just be u=...
To find u, you have to remove what is attached to it, and that is -12. Then you have to look at the relationship between the -12 and u. The relationship is multiplication, and the opposite of multiplication is division, so all you have to do is divide both sides by -12. So;
-12u/-12=-24/-12
The -12 cancels the -12, leaving u and the - in 12 cancels the - in 24. Leaving 24/12. And that is 2. Written as;
u=2
Hope i helped. If you have any more problems, let me know.
<span>, y+2 = (x^2/2) - 2sin(y)
so we are taking the derivative y in respect to x so we have
dy/dx use chain rule on y
so y' = 2x/2 - 2cos(y)*y'
</span><span>Now rearrange it to solve for y'
y' = 2x/2 - 2cos(y)*y'
0 = x - 2cos(y)y' - y'
- x = 2cos(y)y' - y'
-x = y'(2cos(y) - 1)
-x/(2cos(y) - 1) = y'
</span><span>we know when f(2) = 0 so thus y = 0
so when
f'(2) = -2/(2cos(0)-1)
</span><span>2/2 = 1
</span><span>f'(2) = -2/(2cos(0)-1)
cos(0) = 1
thus
f'(2) = -2/(2(1)-1)
= -2/-1
= 2
f'(2) = 2
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