Answer:
<h2>x > 5</h2>
Step-by-step explanation:
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]}
The depth of the water at the <em>deepest</em> point in the waterslide and to the nearest hundredth of a meter is approximately 0.19 meters. (Correct choice: A)
<h3>How to determine the depth of the water in a waterslide</h3>
In this question we should apply the concepts of <em>right</em> triangles and <em>trigonometric</em> functions to determine the height of the water within the waterslide. A geometric diagram of the <em>cross</em> section of the waterslide is presented below, which indicates the existence of <em>circular</em> symmetry.
Now we proceed to determine the height of the water:
cos α = (0.5 m/0.75 m)
α ≈ 48.190°
y = (0.75 m) · sin 48.190°
y = 0.559 m
x = 0.75 m - 0.559 m
x = 0.191 m
The depth of the water at the <em>deepest</em> point in the waterslide and to the nearest hundredth of a meter is approximately 0.19 meters. (Correct choice: A)
To learn more on trigonometry: brainly.com/question/22698523
#SPJ1
<u>Corrected Question</u>
These are the given probabilities. The Venn diagram is also attached. are (A) P(A|C) = 2/3
(B) P(C|B) = 8/27
(C) P(A) = 31/59
(D) P(C) = 3/7
(E) P(B|A) = 13/27
Answer:
A and C
Step-by-step explanation:
From the above, only A and C are correct.
