Answer:
see explanation
Step-by-step explanation:
Using the tangent ratio in the right triangle
tan A =
=
=
, then
∠ A =
(
) ≈ 41° ( to the nearest degree )
The sum of the angles in the triangle = 180° , then
∠ B + 41° + 90° = 180°
∠ B + 131° = 180° ( subtract 131° from both sides )
∠ B = 49°
Using Pythagoras' identity in the right triangle
AB² = BC² + AC² = 7² + 8² = 49 + 64 = 113 ( take square root of both sides )
AB =
≈ 10.6 ( to the nearest tenth )
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:
, 9
Step-by-step explanation:
=
÷
= 
=
= 9
Solution:
f(x) = 2x + 3
g(x) = x - 9
f(x) + g(x)
= (2x + 3) + (x - 9)
= 2x + x + 3 - 9
= 3x - 6