Answer:
The first quadrant
Step-by-step explanation:
Aka upper right quadrant
Answer:
20
Step-by-step explanation:
Answer:
y=t−1+ce
−t
where t=tanx.
Given, cos
2
x
dx
dy
+y=tanx
⇒
dx
dy
+ysec
2
x=tanxsec
2
x ....(1)
Here P=sec
2
x⇒∫PdP=∫sec
2
xdx=tanx
∴I.F.=e
tanx
Multiplying (1) by I.F. we get
e
tanx
dx
dy
+e
tanx
ysec
2
x=e
tanx
tanxsec
2
x
Integrating both sides, we get
ye
tanx
=∫e
tanx
.tanxsec
2
xdx
Put tanx=t⇒sec
2
xdx=dt
∴ye
t
=∫te
t
dt=e
t
(t−1)+c
⇒y=t−1+ce
−t
where t=tanx
Answer:
(-4, -8)
Step-by-step explanation:
Use the substitution method. x = -4, so y = (1/2)x - 6 becomes:
y = (1/2)(-4) - 6, or y = -2 - 6, or y = -8.
The solution is (-4, -8).
Find the perpendicular line then find the intersection then find the point
perpendicular lines have slopes that are perpendicular
the slopes multiply bo -1
y=mx+b
m=slope
y=2x-3
2 is slope
2 times what=-1
what=-1/2
the equation is
y-3=-1/2(x-8) or
y=(-1/2)x+7
find intersection
at (4,5)
distance bwetweeen (8,3) and (4,5)
D=
D=
D=
D=

D=2√5
distance= 2√5