P
pLuvia
Guest
You sure you didn't mean e2x?
If so
If so
e4x-4e2x+4=0
(e2x-2)(e2x-2)=0
e2x=2
x=(ln2)/2
(e2x-2)(e2x-2)=0
e2x=2
x=(ln2)/2
Find the equation of the straight line parallel to the line 2x-3y-1=0 and through the midpoint of (1,3) and (-1,9)SoulSearcher said:Question?
So, it seems just about all Year 12 students doing Mathematics and/or its Extension courses drink Calculus when solving graphical equations .... I should have stated not to use calculus..pLuvia said:......2x-3y-1=0
y=2x/3-1/3
y'=2/3
MP=[(1-1)/2,(3+9)/2]
=(0,6)
Hence
y-6=2x/3
2x-3y+18=0
Diffrentiate to the first principle 2x3-16x5 if x=6sinist4 said:post another Q
mica said:oh ok thanks SoulSearcher
So with 1+sqrt{3}tanx+3tan<sup>2</sup>x+3sqrt{3}tan<sup>3</sup>x+...
with a = 1
and r = t2 / t1
= sqrt{3}tanx
= t3 / t2
= 3tan^2x / sqrt{3}tanx
= 3tanx / sqrt{3}
= 3tanx / sqrt{3} * sqrt{3} / sqrt{3}
= sqrt{3}tanx
therefore to find values of x we place r = sqrt{3}tanx within -1 and 1 as such
-1 < sqrt{3}tanx < 1
= -1 / sqrt{3} < tanx < 1 / sqrt{3}
and 1 / sqrt{3} = pie / 6 using the trig triangles
therefore x = pie + pie / 6 to get -1 / sqrt{3}
and x = pie / 6 to get 1 / sqrt{3}
therefore x is any value between 30 to 210 degrees not including.
30 < x < 210
i wasnt quite sure so someone tell me what i have done wrong
I didn't use calculus at allf3nr15 said:So, it seems just about all Year 12 students doing Mathematics and/or its Extension courses drink Calculus when solving graphical equations .... I should have stated not to use calculus..
Too lazy to write m1, means the same thing anyway Except that there was no calculus involveddrynxz said:y'=2/3 <--- use of y'