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
Answer:
pair of pants = $53
Step-by-step explanation:
s = shirts
p = pair of pants
eq = equation
eq 1: 7s + 5p = 517
eq 2: s + p = 89
multiply the equation 2 by 7
eq 3: 7s + 7p = 623
negate equation 1 and add it to the eq 3
7s + 7p = 623
+ -7s - 5p = -517
2p = 106
p = 53
Answer
15.652476
Reason: i used a calculator
Answer:
Step-by-step explanation:
A quadrilateral with 4 same-length sides is a <em>rhombus</em>.
(If the corner angles are also 90°, then the rhombus is also a square. That is not the case here.)
__
Adjacent angles are supplementary, so ...
x° = 180° -110° = 70°
Sides are the same length, so ...
2y = y+5
y = 5 . . . . . subtract y
_____
If part of the picture was cut off, the right side of the figure may be something other than 2y. If so, formulate the equation appropriately and solve for y.
