Using the unit circle, the coordinates of 3π/4 is :
![(-\frac{\sqrt[]{2}}{2},\frac{\sqrt[]{2}}{2})](https://tex.z-dn.net/?f=%28-%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%2C%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%29)
Note that tangent function can be solved using the formula y/x
where x and y are the coordinates of the angle.
This will be :
![\frac{\sqrt[]{2}}{2}\div(-\frac{\sqrt[]{2}}{2})=-1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%5Cdiv%28-%5Cfrac%7B%5Csqrt%5B%5D%7B2%7D%7D%7B2%7D%29%3D-1)
The answer is -1
I don't know kid I'm busy right now
L# = 1 + 3S#
8S# - 2L# = 10
8S# - 2(1 + 3S#) = 10
8S# - 2 + 6S# = 10
14S# = 12
S# = 6/7
L# = 1 + 3(12/14)
L# = 25/7
<h2><em>y
2
+
6
y
−
16
=
(
y
+
8
)
(
y
−
2
)
</em></h2><h2 /><h2><em>Explanation:
</em></h2><h2><em>Note that in general:
</em></h2><h2 /><h2><em>(
y
+
a
)
(y
−
b
)
=
y
2
+
(
a
−
b
)
y
−
a
b
</em></h2><h2 /><h2><em>So we want to find a pair of factors a and b of 16 which differ by 6
. </em></h2><h2><em>The pair 8
,
2 works in that 8−
2
=
6 and 8
⋅
2=
16
.
</em></h2><h2 /><h2><em>Hence:
</em></h2><h2><em>
y
2
+
6
y
−
16
=
(
y+
8
)
(
y
−
2
)</em></h2><h2><em>y=2,-8</em></h2>
