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
i mean ig but 129.... ehh
Answer:
y intercept (0, 4)
x intercept (4, 0)
I hope this is good enough:
Since the slope is -2 because
x1: -1 x2: 0
y1: 2 y2: 0
y2-y1/x2-x1 = slope (0-2/0+1= -2/1 = -2)
You would havr to change it into point slope form so it would look like this
Y-2 = -2(x+1)
y-2 = -2x -2
+2 +2
y = -2x + 0
Hope this helps:)
Answer: x = 83 (x representing the next homework assignment)
Step-by-step explanation:
