Answer:
D. People said that man could not fly, but in 1903, Orville and Wilbur Wright made the first sustained powered flight in a heavier-than-air machine.
Step-by-step explanation:
Step-by-step explanation:
Move expression to the left side and change its sign
5
y
−
3
+
10
y
2
−
y
−
6
−
y
y
+
2
=
0
Write
−
y
as a sum or difference
5
y
−
3
+
10
y
2
+
2
y
−
3
y
−
6
−
y
y
+
2
=
0
Factor out
y
and
−
3
from the expression
5
y
−
3
+
10
y
(
y
+
2
)
−
3
(
y
+
2
)
−
y
y
+
2
=
0
Factor out
y
+
2
from the expression
5
y
−
3
+
10
(
y
+
2
)
(
y
−
3
)
−
y
y
+
2
=
0
Write all numerators above the least common denominator
5
(
y
+
2
)
+
10
−
y
(
y
−
3
)
(
y
+
2
)
(
y
−
3
)
=
0
Distribute
5
and
−
y
through the parenthesis
5
y
+
10
+
10
−
y
2
+
3
y
(
y
+
2
)
(
y
−
3
)
=
0
Collect the like terms
8
y
+
20
−
y
2
(
y
+
2
)
(
y
−
3
)
=
0
Use the commutative property to reorder the terms
−
y
2
+
8
y
+
20
(
y
+
2
)
(
y
−
3
)
=
0
Write
8
y
as a sum or difference
−
y
2
+
10
y
−
2
y
+
20
(
y
+
2
)
(
y
−
3
)
=
0
Factor out
−
y
and
−
2
from the expression
−
y
(
y
−
10
)
−
2
(
y
−
10
)
(
y
+
2
)
(
y
−
3
)
=
0
Factor out
−
(
y
−
10
)
from the expression
−
(
y
−
10
)
(
y
+
2
)
(
y
+
2
)
(
y
−
3
)
=
0
Reduce the fraction with
y
+
2
−
y
−
10
y
−
3
=
0
Determine the sign of the fraction
−
y
−
10
y
−
3
=
0
Simplify
10
−
y
y
−
3
=
0
When the quotient of expressions equals
0
, the numerator has to be
0
10
−
y
=
0
Move the constant,
10
, to the right side and change its sign
−
y
=
−
10
Change the signs on both sides of the equation
y
=
10
Check if the solution is in the defined range
y
=
10
,
y
≠
3
,
y
≠
−
2
∴
y
=
10
When looking at expressions like this,, you need to remember that the square root of a negative number does not exist in the set of Real Numbers. This means that there is no way to simplify this expression,, making the correct answer to your question Undefined.
Let me know if you have any further questions.
:)
and 
. The squared distance between any point in this region 
and the point (1, 1) is
![=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7D%5Br%5E2-2r%28%5Ccos%5Ctheta-%5Csin%5Ctheta%29%2B2%5Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D)
