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
Three loaves since that’s what’s available to the family
Answer:
You could use the polygon area formula OR here's an easier solution
We draw the lines DC and DE
Angle ADC = 120°
Triangle ADC is a 30 60 90 triangle
In such a triangle Line AE = hypotenuse * sq root (3) /2
Line AE = 3 * 0.8660254038 = 2.5980762114
Line DE = 1.5
Area of Triangle ADE = .5 * 1.5 * 2.5980762114 =
1.9485571586 square meters and entire area of Triangle
ABC = 6 * 1.9485571586
which equals 11.6913429513 square meters
Step-by-step explanation:
Answer:
14.2
Step-by-step explanation: