If you take the derivative of your equation, you get:
2′″−″−′+′=0
2
y
′
y
″
−
x
y
″
−
y
′
+
y
′
=
0
or
″(2′−)=0.
y
″
(
2
y
′
−
x
)
=
0.
Let =′
v
=
y
′
and we have ′(2−)=0,
v
′
(
2
v
−
x
)
=
0
,
so either ′=0
v
′
=
0
and =
v
=
c
or =/2
v
=
x
/
2
.
Then ′=
y
′
=
c
and so =+
y
=
c
x
+
d
or ′=/2
y
′
=
x
/
2
and =2/4.
y
=
x
2
/
4.
Plugging the first into the original equation gives =−2
d
=
−
c
2
. So there are two solutions =−2
y
=
c
x
−
c
2
for some constant
c
and =2/4
y
=
x
2
/
4
. I don't know if this is all the solutions
(Square)
Base length-6 units
Height-4 units
Volume-48 cubic units
Volume of 4 square pyramids-192 cubic units
(Rectangular)
Base length-12 units
Base width-8 units
Height-6 units
Volume-192 cubic units
(What I got. Though the answers vary because there's a bunch of ways to do it. This is just one. Hope it helps)
Answer:
(A) (-5)
Step-by-step explanation:
B and C are even
D (-9) is less than (-7)
-1 plus 5 is 4 so the answer is a
Answer:
14 each
Step-by-step explanation:
33-5=29
28/2=14
14 each
