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:
Step-by-step explanation:
![\sqrt[3]{125y^9z^6}\\ \\ \sqrt[3]{5^3(y^3)^3(z^2)^3}\\ \\ 5y^3z^2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125y%5E9z%5E6%7D%5C%5C%20%5C%5C%20%5Csqrt%5B3%5D%7B5%5E3%28y%5E3%29%5E3%28z%5E2%29%5E3%7D%5C%5C%20%5C%5C%205y%5E3z%5E2)
The answer is option c.
That is, the wrong step in step 6. It was written that the center of the circunference is the point (2.1). However, the general equation of a circumference is:
(X- (a)) ^ 2 + (Y- (b)) ^ 2 = r ^ 2
Where the point (a, b) is the center of the circle.
So for this case the point for the center is: (-2, -1)
Answer:
the answer is 9
Step-by-step explanation:
15/100 ?/60
multiply 60 by 15 to get 900, the divide 900 by 100 to get <em>9</em>
