To start, you can use clue 2 because there is only one variable in it.
Solve for the value of the circle, by dividing both sides by 3.
Then you can plug that value into clue 1 to solve for the diamond.
Once you find the value for the diamond, you can plug both values, the diamond and the circle, into clue 3, to solve for the triangle.
Answer:
-2k
Step-by-step explanation:
-k +3k
Factor out k
k(1-3)
k (-2)
-2k
Answer:
Explanation:
An isosceles triangle is one with two equal lengths.
To find the lengths of the sides with coordinates
(
x
1
,
y
1
)
,
(
x
2
,
y
2
)
we use Pythagoras' theorem
d
=
√
(
x
2
−
x
2
)
2
+
(
y
2
−
y
1
)
2
from the information given
d
1
=
√
(
13
−
9
)
2
+
(
−
2
−
−
8
)
2
d
1
=
√
4
2
+
6
2
=
√
52
−
−
(
1
)
d
2
=
√
(
13
−
5
)
2
+
(
−
2
−
−
2
)
2
d
2
=
√
8
2
+
0
2
=
√
64
=
8
−
−
(
2
)
d
3
=
√
(
9
−
5
)
2
+
(
−
8
−
−
2
)
2
d
3
=
√
4
2
+
6
2
=
√
52
−
−
−
(
3
)
we see
d
1
=
d
3Explanation:
An isosceles triangle is one with two equal lengths.
To find the lengths of the sides with coordinates
(
x
1
,
y
1
)
,
(
x
2
,
y
2
)
we use Pythagoras' theorem
d
=
√
(
x
2
−
x
2
)
2
+
(
y
2
−
y
1
)
2
from the information given
d
1
=
√
(
13
−
9
)
2
+
(
−
2
−
−
8
)
2
d
1
=
√
4
2
+
6
2
=
√
52
−
−
(
1
)
d
2
=
√
(
13
−
5
)
2
+
(
−
2
−
−
2
)
2
d
2
=
√
8
2
+
0
2
=
√
64
=
8
−
−
(
2
)
d
3
=
√
(
9
−
5
)
2
+
(
−
8
−
−
2
)
2
d
3
=
√
4
2
+
6
2
=
√
52
−
−
−
(
3
)
we see
d
1
=
d
3Explanation:
An isosceles triangle is one with two equal lengths.
To find the lengths of the sides with coordinates
(
x
1
,
y
1
)
,
(
x
2
,
y
2
)
we use Pythagoras' theorem
d
=
√
(
x
2
−
x
2
)
2
+
(
y
2
−
y
1
)
2
from the information given
d
1
=
√
(
13
−
9
)
2
+
(
−
2
−
−
8
)
2
d
1
=
√
4
2
+
6
2
=
√
52
−
−
(
1
)
d
2
=
√
(
13
−
5
)
2
+
(
−
2
−
−
2
)
2
d
2
=
√
8
2
+
0
2
=
√
64
=
8
−
−
(
2
)
d
3
=
√
(
9
−
5
)
2
+
(
−
8
−
−
2
)
2
d
3
=
√
4
2
+
6
2
=
√
52
−
−
−
(
3
)
we see
d
1
=
d
3
1/4 - 1/6 = (6-4)/24 = 2/24 = 1/12
1/3 - 1/4 = (4-3)/12 = 1/12
Answer:
177.6
Step-by-step explanation: