Answer: the rectangle of maximum area is a square of side length 1= 4 cm let x and y be the lengths of the sides of the rectangle measured in cm
Step-by-step explanation:
then the perimeter is
2x+2y=16
so that:
y=8- x
Step-by-step explanation:
1/2 (c +1)>5 = c≥9
27-2b<-6=b≥332
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
Graph using y=mx+b. The point of intersection is (3,8)
Answer:
D
Step-by-step explanation:
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