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
7x^2 - 16x - 8 = 0
using the quadratic formula x = [ -(-16)+/- sqrt((-16^2 - 4*7-8] / 2*7
= (16 +/1 sqrt 480) / 14
= 2.71, -0.42 to nearest hundredth
Answer:
<h2>231cm²</h2>
Step-by-step explanation:
First, let's find the surface area of both the triangles
5x3=15
So, the surface area of the triangles is 15 sq.cm
Now, let's find the surface area of the base (large rectangle in the middle)
12x8=?
10x8=80
2x8=16
80+16=96
12x8=96
So, the surface area of the base, is 96sq.cm
Now, let's find the surface area of both of the side rectangles
12x5=60
60x2=120
So, the surface area of the two side rectangles is 120sq.cm
Now, let's find the total surface area by adding all of our answers.
120+96=216
216+15=231
<h2>
So hence, the surface area of this net is 231cm²</h2>
Answer:
The answer for this is x=-5
