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:
Class 8 has 27 and class 7 has 40
Step-by-step explanation:
8th (r) class has 13 fewer than 7th (x).
x -13 = r
Total for both classes = 67
r + x - 13 = 67
r + x = 80
80 divided by two = 40 (per class)
40 - 13 = 27
Since ΔABC ~ ΔEDC, ∠B = ∠D.
Since both triangles appear to be similar, the corresponding angles are the same, and corresponding sides are the same or have the same ratio.
We can write an equation to resemble the problem:
8x + 16 = 120
Solve for x.
8x + 16 = 120
~Subtract 16 to both sides
8x + 16 - 16 = 120 - 16
~Simplify
8x = 104
~Divide 8 to both sides
8x/8 = 104/8
~Simplify
x = 13
Therefore, the answer is 13.
Best of Luck!
Answer:
1
Step-by-step explanation:
Any number raised to the 0th power is 1, except for 0; 0 to the 0th power is undefined.
Answer:
x int (4,0)
y int (0, -6)
Step-by-step explanation:
