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:
it is
Step-by-step explanation:
Answer:
(0,-6), (-2,0)
Step-by-step explanation:
-3x-y=6
-y=6+3x
y=-3x-6
when x=0, y=-6
when y=0 x=-2
Answer:
16 bombons de coco
Step-by-step explanation:
A partir da pergunta, somos instruídos a fingir que o número total de bombons (chocolate) que ela fez = 24 bombons
Do total de chocolates que ele fez, 2/3 são de coco.
Portanto, a quantidade que representa os doces de coco é representada como:
2/3 × 24
= 16
Portanto, a quantidade de bombons de coco = 16
