Answer:
21
20.7
20.23
20.22
20.102
20.005
Step-by-step explanation:
hope this helps!
You can always find the area from breaking it up into smaller shapes. For example, a hexagon is basically 2 trapezoids.
Some formulas for different shapes.
Trapezoid: <span>height × (base1 + base2) / 2
Rectangle: length x width
Parallelogram: height x base
Triangle height x base / 2</span>
Answer:
there are 25 students in the class and 15 who wear glasses.
Was there a picture or must you write a visual graph? It should like a small increase in heart rate then a steady none moving line and then slowly decreasing downward to nothing. :)
Answer:
A - Rectangle B - Square
C - Parallelogram D - Rhombus
Explanation:
We are given
A
(
1
,
2
)
,
B
(
2
,
−
2
)
and hence
A
B
=
√
(
2
−
1
)
2
+
(
−
2
−
2
)
2
=
√
17
. Further slope of
A
B
is
−
2
−
2
2
−
1
=
−
4
1
=
−
4
.
Case A -
C
(
−
6
,
−
4
)
,
D
(
−
7
,
0
)
As
C
D
=
√
(
−
7
−
(
−
6
)
)
2
+
(
0
−
(
−
4
)
)
2
=
√
17
and slope of
C
D
is
0
−
(
−
4
)
−
7
−
(
−
6
)
=
4
−
1
=
−
4
As
A
B
=
C
D
and
A
B
||
C
D
slopes being equal, ABCD is a parallelogram.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x+6)^2+(y+4)^2-0.08)((x+7)^2+y^2-0.08)=0 [-10, 10, -5, 5]}
Case B -
C
(
6
,
−
1
)
,
D
(
5
,
3
)
As
C
D
=
√
(
5
−
6
)
2
+
(
3
−
(
−
1
)
)
2
=
√
17
and slope of
C
D
is
0
−
(
−
4
)
−
7
−
(
−
6
)
=
4
−
1
=
−
4
Further,
B
C
=
√
(
6
−
2
)
2
+
(
−
1
−
(
−
2
)
)
2
=
√
17
and slope of
B
C
is
−
1
−
(
−
2
)
6
−
2
=
1
4
As
B
C
=
A
B
and they are perpendicular (as product of slopes is
−
1
), ABCD is a square.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x-6)^2+(y+1)^2-0.08)((x-5)^2+(y-3)^2-0.08)=0 [-10, 10, -5, 5]}
Case C -
C
(
−
1
,
−
4
)
,
D
(
−
2
,
0
)
As mid point of
A
C
is
(
1
−
1
2
,
2
−
4
2
)
i.e.
(
0
,
−
1
)
and midpoint of
B
D
is
(
2
−
2
2
,
−
2
+
0
2
i.e.
(
0
,
−
1
)
i.e. midpoints of
A
C
and
B
D
are same,
but,
B
C
=
√
(
2
−
(
−
1
)
)
2
+
(
−
2
−
(
−
4
)
)
2
=
√
13
i.e.
A
B
≠
B
C
and hence ABCD is a parallelogram.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x+1)^2+(y+4)^2-0.08)((x+2)^2+y^2-0.08)=0 [-10, 10, -5, 5]}
Case D -
C
(
1
,
−
6
)
,
D
(
0
,
−
2
)
As mid point of
A
C
is
(
1
+
1
2
,
2
−
6
2
)
i.e.
(
1
,
−
2
)
and midpoint of
B
D
is
(
2
+
0
2
,
−
2
+
(
−
2
)
2
i.e.
(
1
,
−
2
)
i.e. midpoints of
A
C
and
B
D
are same,
and,
B
C
=
√
(
2
−
1
)
2
+
(
−
2
−
(
−
6
)
)
2
=
√
17
i.e.
A
B
=
B
C
and hence ABCD is a rhombus.
graph{((x-1)^2+(y-2)^2-0.08)((x-2)^2+(y+2)^2-0.08)((x-1)^2+(y+6)^2-0.08)(x^2+(y+2)^2-0.08)=0 [-14, 14, -7, 7]}
