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 2 equations are
18.20x+19.50y=230.10
and
x+y=12
where x is the months of original cost and y is months for new cost. Since you know that you paid for one year (12 months) you can make the second equation. Then you want to substitute the first equations x by making the second equation
x=(12-y)
18.20(12-y)+19.50y=230.10
218.40-18.20y+19.50y=230.10
1.30y=11.70
y=9
so that means you had the original rate for 3 months and the new one for 9 months
Answer:
(x−5y)(9x+8)
Step-by-step explanation:
9x^2−45yx+8x−40y
Do the grouping 9x^2−45yx+8x−40y=(9x^2−45yx)+(8x−40y), and factor out 9x in the first and 8 in the second group.
9x(x−5y)+8(x−5y)
Factor out common term x−5y by using distributive property.
(x−5y)(9x+8)
XY=10
Y=X+3
X(X+3)=10
X^2+3X=10
X^2+3X-10
(X-2)(X+5)
