Answer: 3
x
−
2
y
−
15
=
0
Explanation:
We know that,
the slope of the line
a
x
+
b
y
+
c
=
0
is
m
=
−
a
b
∴
The slope of the line
2
x
+
3
y
=
9
is
m
1
=
−
2
3
∴
The slope of the line perpendicular to
2
x
+
3
y
=
9
is
m
2
=
−
1
m
1
=
−
1
−
2
3
=
3
2
.
Hence,the equn.of line passing through
(
3
,
−
3
)
and
m
2
=
3
2
is
y
−
(
−
3
)
=
3
2
(
x
−
3
)
y
+
3
=
3
2
(
x
−
3
)
⇒
2
y
+
6
=
3
x
−
9
⇒
3
x
−
2
y
−
15
=
0
Note:
The equn.of line passing through
A
(
x
1
,
y
1
)
and
with slope
m
is
y
−
y
1
=
m
(
x
−
x
1
)3
x
−
2
y
−
15
=
0
Explanation:
We know that,
the slope of the line
a
x
+
b
y
+
c
=
0
is
m
=
−
a
b
∴
The slope of the line
2
x
+
3
y
=
9
is
m
1
=
−
2
3
∴
The slope of the line perpendicular to
2
x
+
3
y
=
9
is
m
2
=
−
1
m
1
=
−
1
−
2
3
=
3
2
.
Hence,the equn.of line passing through
(
3
,
−
3
)
and
m
2
=
3
2
is
y
−
(
−
3
)
=
3
2
(
x
−
3
)
y
+
3
=
3
2
(
x
−
3
)
⇒
2
y
+
6
=
3
x
−
9
⇒
3
x
−
2
y
−
15
=
0
Note:
The equn.of line passing through
A
(
x
1
,
y
1
)
and
with slope
m
is
y
−
y
1
=
m
(
x
−
Explanation:
the equation of a line in
slope-intercept form
is.
∙
x
y
=
m
x
+
b
where m is the slope and b the y-intercept
rearrange
2
x
+
3
y
=
9
into this form
⇒
3
y
=
−
2
x
+
9
⇒
y
=
−
2
3
x
+
3
←
in slope-intercept form
with slope m
=
−
2
3
Given a line with slope then the slope of a line
perpendicular to it is
∙
x
m
perpendicular
=
−
1
m
⇒
m
perpendicular
=
−
1
−
2
3
=
3
2
⇒
y
=
3
2
x
+
b
←
is the partial equation
to find b substitute
(
3
,
−
3
)
into the partial equation
−
3
=
9
2
+
b
⇒
b
=
−
6
2
−
9
2
=
−
15
2
⇒
y
=
3
2
x
−
15
2
←
equation of perpendicular lineExplanation:
the equation of a line in
slope-intercept form
is.
∙
x
y
=
m
x
+
b
where m is the slope and b the y-intercept
rearrange
2
x
+
3
y
=
9
into this form
⇒
3
y
=
−
2
x
+
9
⇒
y
=
−
2
3
x
+
3
←
in slope-intercept form
with slope m
=
−
2
3
Given a line with slope then the slope of a line
perpendicular to it is
∙
x
m
perpendicular
=
−
1
m
⇒
m
perpendicular
=
−
1
−
2
3
=
3
2
⇒
y
=
3
2
x
+
b
←
is the partial equation
to find b substitute
(
3
,
−
3
)
into the partial equation
−
3
=
9
2
+
b
⇒
b
=
−
6
2
−
9
2
=
−
15
2
⇒
y
=
3
2
x
−
15
2
←
equation of perpendicular line
The answer is C, Region A.
Answer:
A.
Step-by-step explanation:
EF^2=18^2-12^2
EF^2=324-144
EF^2=180
EF=root 180
Answer:
The solutions of the system of equations are the points
Step-by-step explanation:
we have
----> equation A
----> equation B
Solve the system by substitution
substitute equation B in equation A
solve for x
The formula to solve a quadratic equation of the form
is equal to
in this problem we have
so
substitute in the formula
-----> 
-----> 
<em>Find the values of y</em>
For
---->
For
---->
therefore
The solutions of the system of equations are the points
Answer:
6xy²z^4
Step-by-step explanation:
![\sqrt[3]{216x^3y^6z^12}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B216x%5E3y%5E6z%5E12%7D)
= ∛(6³)(x³)(y³)(y³)(z³)(z³)(z³)(z³) = 6x·y·y·z·z·z·z = 6xy²z^4
