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Ask QuestionPosted by Samyak Jain 6 years, 11 months ago
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Posted by Surabhi Agarwal 6 years, 11 months ago
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Shubhendra Singh ? 6 years, 11 months ago
Posted by Saurabh Gangwar 6 years, 11 months ago
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Posted by Avinash Raghuvanshi 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
First, we form a table for all three equations different value of x and y, As:
for, y = x
| x | 1 | 2 | 3 |
| y | 1 | 2 | 3 |
for, x + y = 6
| x | 6 | 2 | 3 |
| y | 0 | 4 | 3 |
for, y = 2x
| x | 1 | 2 | 3 |
| y | 2 | 4 | 6 |
Now we represent these points on the X-Y plane. As:
The area of the shaded region
{tex}= A ( \Delta A B C ) - A ( \Delta A D C ){/tex}
{tex}= \frac { 1 } { 2 } \times \text { height of } \Delta A B C \times A C + \frac { 1 } { 2 } \times \text { height of } \Delta A D C \times A C{/tex}
{tex}= \frac { 1 } { 2 } \times 4 \times 6 + \frac { 1 } { 2 } \times 3 \times 6{/tex}
= 21 cm2
Posted by Rajkamal Patel 6 years, 11 months ago
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Posted by Pooja Nangliya 6 years, 11 months ago
- 4 answers
Chētnà Pandey✌️ 6 years, 11 months ago
Chētnà Pandey✌️ 6 years, 11 months ago
Posted by Pooja Nangliya 6 years, 11 months ago
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Prerna Baid 6 years, 11 months ago
Puja Sahoo 6 years, 11 months ago
Posted by Pooja Nangliya 6 years, 11 months ago
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Posted by Pooja Kumari 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
| 11-13 | 3 | 12 | 36 |
| 13-15 | 6 | 14 | 84 |
| 15-17 | 9 | 16 | 144 |
| 17-19 | 13 | 18 | 234 |
| 19-21 | f | 20 | 20f |
| 21-23 | 5 | 22 | 110 |
| 23-25 | 4 | 24 | 96 |
| {tex}\sum f _ { i } = 40 + f{/tex} | {tex}\sum f _ { i } x _ { i } = 704 + 20 f{/tex} |
let the missing frequency is 'f'.
we know that , {tex}Mean= \frac { \sum f _ { i } x _ { i } } { \sum f _ { i } }{/tex}
{tex}\Rightarrow 18 = \frac { 704 + 20 f } { 40 + f }{/tex}
{tex}\Rightarrow 18(40 + f)=704 + 20f{/tex}
{tex}\Rightarrow 720 + 18f =704 + 20f{/tex}
{tex}\Rightarrow 2f = 16{/tex}
{tex}\Rightarrow{/tex} f = 8
Posted by Kshitiz Narayan 6 years, 11 months ago
- 1 answers
Yogita Ingle 6 years, 11 months ago
In geometry, a frustum is the portion of a solid (normally a cone or pyramid) that lies between one or two parallel planes cutting it.
If we cut a right circular cone by a plane parallel to its base, the portion of the solid between this plane and the base is known as the frustum of a cone.
Curved Surface Area (CSA) = pi * l(R + r)
where
r = radius of smaller circle
R = radius of bigger circle
l = slant height of the frustum
Posted by Anshika Dixit 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let {tex}x = \sqrt { 6 + \sqrt { 6 + \sqrt { 6 + \dots } } }{/tex} ......(i)
{tex}\Rightarrow{/tex} x2 = {tex}( \sqrt { 6 + \sqrt { 6 + \sqrt { 6 . . } } } ) ^ { 2 }{/tex} [Squaring both sides]
{tex}\Rightarrow{/tex} x2 = {tex}6 + \sqrt { 6 + \sqrt { 6 + \sqrt { 6 + \ldots } } }{/tex}
{tex}\Rightarrow{/tex} {tex}x^2 = 6 + x{/tex} [From (i)]
{tex}\Rightarrow{/tex} {tex}x^2 - x - 6 = 0{/tex}
{tex}\Rightarrow{/tex} {tex}(x - 3) (x + 2) = 0{/tex}
{tex}\Rightarrow{/tex} {tex}x = 3, x = -2{/tex}
{tex}\therefore{/tex} x = 3 [{tex}\because{/tex} x > 0]
Posted by Harsimar Singh 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Given
Volume of cistern = {tex}150 \times 120 \times 110 \mathrm { cm } ^ { 3 } = 1980000 \mathrm { cm } ^ { 3 }{/tex}
Volume of water = 129600 cm3
Volume of one brick = {tex}22.5 \times 7.5 \times 6.5 \mathrm { cm } ^ { 3 } = 1096.875 \mathrm { cm } ^ { 3 }{/tex}
Each brick absorbs one - seventeenth of its volume of water
Volume of water absorbed by one brick = {tex}\frac { 1 } { 17 } \times{/tex} volume of brick
= {tex}\frac { 1 } { 17 } \times 1096.875 \mathrm { cm } ^ { 3 }{/tex}
= 64.52 cm3
Let n be the total number of bricks which can be put in the cistern without water overflowing. Then,
Volume of water absorbed by n bricks = {tex}n \times \frac { 1 } { 17 } \times 1096.875 \mathrm { cm } ^ { 3 }{/tex}
{tex}\therefore{/tex} Volume of water left in cistern = {tex}= \left( 129600 - \frac { n } { 17 } \times 1096.875 \right) \mathrm { cm } ^ { 3 }{/tex}
Since the cistern is filled upto the brim.
Therefore, Volume of the cistern = Volume of water left in the cistern + Volume of bricks
{tex} 1980000{/tex} = {tex}129600 - \frac { n } { 17 } \times 1096.875 + n \times 1096.875 {/tex}
{tex}n \times 1096.875 - \frac { n } { 17 } \times 1096.875 = 1980000 - 129600{/tex}
{tex}1096.875 \times \left( n - \frac { n } { 17 } \right) = 1850400{/tex}
{tex}1096.875 \times \frac { 16 n } { 17 } = 1850400{/tex}
{tex}17550 \times \frac { n } { 17 } = 1850400 \Rightarrow n = \frac { 1850400 \times 17 } { 17550 } = 1792.41 {/tex}
since the number of bricks cannot be in decimals
therefore, required number of bricks = 1792
Posted by Ketan Pandey 6 years, 11 months ago
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Posted by Anupam Kumar 6 years, 11 months ago
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Posted by Shoeb Akhtar 6 years, 11 months ago
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Ram Kushwah 6 years, 11 months ago
A line segment is a piece, or part, of a line in geometry. A line segment is represented by end points on each end of the line segment. A line in geometry is represented by a line with arrows at each end. ... For example, if your end points were A and B, then you would write your line segment AB with a line over the top.
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Abc . 6 years, 11 months ago
Shubhendra Singh ? 6 years, 11 months ago
Neha Jangir 6 years, 11 months ago
Mtror Thardak 6 years, 11 months ago
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Aryan Beniwal 6 years, 11 months ago
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Roshan Raj 6 years, 11 months ago
1Thank You