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
-2x + 6 = 30 - 6x
--> -2x + 6x = 30 - 6
<span>--> 4x = 24
</span><span>--> x = 24/4
</span><span>--> x = 6</span>
Your distance is 1424 and your rate is 356. As you said, d = r*t. Since you don't have the time you would do 1424 divided by 356. This leaves you with a time of 4 hours. (Don't forget units!)
The weight of a 170 cm steel bar will be 5 Kg
Step-by-step explanation:
Derek uses a 136 cm flat steel bar that weighs 4 kg to make rack in the garage.
1 kg = 1000 gm
So the weight of 1 cm steel bar will be
kg
Weight of 1 cm bar =
gm
Let the weight of a 170 cm steel bar will be
×
gm
⇒
×
gm
⇒ 5000 gm
⇒5 kg
Hence, the weight of a 170 cm steel bar will be 5 Kg
Answer:
i think light blue
Step-by-step explanation:
that or dark blue but you have to graph the answer to find where it lands.