Answer:
Hence the new height is 3 times the original height .(h1=3h)
Step-by-step explanation:
Given:
A cone with has height and base with radius r .
To Find:
What is new height
Solution:
Consider as cone with height h base radius r, and volume v
Here given that only height changes for the cone i.e. r remains the unchanged or same or constant
The volume for a regular cone is given by ,
Here V is directly proportional to h i.ee pie ,3 and r being constant
i.e V/h=constant
V1 and h1 are new dimensions for new cone
V/h=V1/h1
Here V1=3V
So V/h=3V/h1
1/h=3/h1
i.e h1=3h
Hence the height is 3 times the original height .
Answer:
Deepak: 4 t-shirts,
Tarun: 12 t-shirts,
Mahesh: 14 t-shirts.
Step-by-step explanation:
Let x represent number of t-shirts that Deepak has.
Please consider the complete question.
Tarun has 4 t-shirt more than twice the number of T-shirts Deepak has. Mahesh has 2 more than thrice the number of T-shirts Deepak has. If the ratio of the t-shirt that Tarun and Mahesh have is 6 : 7, find out the number of the t-shirt each of them has.
Since Tarun has 4 t-shirt more than twice the number of T-shirts Deepak has, so the number of t-shirts that Tarun has would be 
.
We are also told that Mahesh has 2 more than thrice the number of T-shirts Deepak has. So the number of t-shirts that Mahesh has would be 
.
Since the ratio of the t-shirt that Tarun and Mahesh have is 6 : 7, so we can represent this information in an equation as:
Cross multiply:
Therefore, Deepak has 4 t-shirts.
The number of t-shirts that Tarun has would be 
Therefore, Tarun has 12 t-shirts.
The number of t-shirts that Mahesh has would be 
Therefore, Mahesh has 14 t-shirts.
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 ANSWER IS(<span><span><span>w+6)</span>2</span><span><span>(w+6)</span>2</span></span>
Solve equations ,4 ≤ -4x+28, x>= -6
