Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Smita Jaiswal 6 years, 6 months ago
- 0 answers
Posted by Diya Garg 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Given the diameter = 10 m
So, the radius of the well = 5 m
Height of the well = 14 m
Width of the embankment = 5m
Therefore, radius of the embankment = 5 + 5 =10 m
Let h' be the height of the embankment,
Hence, the volume of the embankment = Volume of the well
That is, {tex}\pi{/tex}(R - r)2h' = {tex}\pi r ^ { 2 } h{/tex}
{tex}\Rightarrow{/tex} (102 - 52) {tex}\times{/tex} h' = 52 {tex}\times{/tex} 14
{tex}\Rightarrow{/tex} (100 - 25) {tex}\times{/tex} h' = 25 {tex}\times{/tex} 14
{tex}\Rightarrow h' = \frac { 25 \times 14 } { 75 } = \frac { 14 } { 3 }{/tex}
Therefore, h' = 4.67 cm approximately.
Posted by Diya Garg 6 years, 6 months ago
- 1 answers
Posted by Diya Garg 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Suppose r be the common radius of a cylinder, cone and a sphere.
height of the cylinder = Height of the cone = Height of the sphere {tex}= 2r{/tex}
Let 'l’ be the slant height of the cone. Then
{tex}l ={/tex} {tex}\sqrt{r^2 + h^2}{/tex}= {tex}\sqrt{r^2 + (2r)^2}{/tex} = {tex}\sqrt{5}{/tex}r
{tex}S_1 ={/tex} Curved surface area of cylinder {tex}= 2\pi rh{/tex}
{tex}= 2\pi r . 2r = 4\pi r^2{/tex}
{tex}S_2 ={/tex} Curved surface area of cone = {tex}\pi{/tex}{tex}rl ={/tex} {tex}\pi{/tex}r{tex}\sqrt{5}{/tex}{tex}r = {/tex}{tex}\sqrt{5}{/tex}{tex}\pi{/tex}{tex}r^2 {/tex}
{tex}S_3={/tex} Curved surface area of sphere {tex}= 4\pi r^2{/tex}
{tex}S_1 : S_2 : S_3 = 4\pi r^2 :\sqrt5\pi r^2 : 4\pi r^2{/tex}
{tex}\therefore{/tex}{tex}S_1 : S_2 : S_3 = 4 :\sqrt5 : 4{/tex}
Posted by Soni Jain 6 years, 6 months ago
- 0 answers
Posted by Aryan Raj 6 years, 6 months ago
- 1 answers
Posted by Vinayak Pradhan 6 years, 6 months ago
- 0 answers
Posted by V Siva 6 years, 6 months ago
- 2 answers
Sia ? 6 years, 6 months ago
x + 1/x = 7
We know the algebraic identity,
a³ + b³ + 3ab ( a + b ) = ( a + b )³
or
a³ + b³ = ( a + b )³ - 3ab( a + b )
x³ + 1/x³ = (x + 1/x )³- 3 × x ×1/x(x + 1/x )
= ( x + 1/x )³ - 3( x + 1/x )
= 7³ - 3 × 7
= 7( 7² - 3 )
= 7 ( 49 - 3 )
= 7 × 46
= 322
Posted by Arshpreet Kaur 6 years, 6 months ago
- 3 answers
Posted by Krish Oberoi 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
{tex}\Rightarrow x ^ { 2 } + 3 \sqrt { 2 } x + 4{/tex}
{tex}\Rightarrow x ^ { 2 } + 2 \sqrt { 2 } x + \sqrt { 2 } x + 4{/tex}
{tex}= x ( x + 2 \sqrt { 2 } ) + \sqrt { 2 } ( x + 2 \sqrt { 2 } ){/tex}
{tex}\Rightarrow ( x + 2 \sqrt { 2 } ) \quad ( x + \sqrt { 2 } ){/tex}
{tex}\{ x = - 2 \sqrt { 2 } , - \sqrt { 2 } \}{/tex}
Posted by Aastha Soni 6 years, 6 months ago
- 2 answers
Yogita Ingle 6 years, 6 months ago
Sides Of a Triangular Field are 50 cm , 45 cm, 35 cm
Let, First Side = a
Second Side = b
Third Side = c
We know that , Semi-perimeter = a+b+c/2
⇒ s = 50+45+35/2
⇒ = 50+80/2
⇒ = 130/2
⇒ = 65
Hence semi-perimeter is 65
Using Heron's Formula
Area = √s(s-a)(s-b)(s-c)
⇒ = √65(65-50)(65-45)(65-35)
⇒ = √65(15)(20)(30)
⇒ = √ 585000
⇒ area = 764.85 cm²
Now, we have to find
Required number of flower beds that can be prepared if each bed measures 5m²
=> 764.85m² /5m²
Sia ? 6 years, 6 months ago
s = (a+b+c)/2
where a, b, and c are the length of the three sides of a triangle
s = (35 + 45 + 50) / 2
= 130 / 2
= 65 cm
Area of scalene triangle is given by
A = sqrt (s (s - a) (s - b) (s - c))
= sqrt (65 * (65 - 35) ( 65 - 45) (65 - 50))
= sqrt (65 * 30 * 20 * 15)
= sqrt (585000)
= 764.853 sq cm
Posted by Shikhar Rai 6 years, 6 months ago
- 0 answers
Posted by Akash Verma 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
6x2 + 17x + 5
= 6x2 + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1)(2x + 5)
Posted by Pratham Verma 6 years, 6 months ago
- 0 answers
Posted by Sharannya Srivastav 6 years, 6 months ago
- 1 answers
Posted by Ran3E Genius 6 years, 6 months ago
- 2 answers
Posted by Aishik Das 6 years, 6 months ago
- 0 answers
Posted by Abhishek Kumar 6 years, 6 months ago
- 3 answers
Tanmaya Chita Padhi 6 years, 6 months ago
Posted by Amritesh Raj 6 years, 6 months ago
- 4 answers
Posted by Anupam Dev 6 years, 6 months ago
- 0 answers
Posted by Maya Choudhary 6 years, 6 months ago
- 2 answers
Posted by Yervoda Gauttam 6 years, 6 months ago
- 0 answers
Posted by Jayant Narang 6 years, 6 months ago
- 1 answers
Posted by Aryan Raj 6 years, 6 months ago
- 3 answers
Posted by Kumar Vivek 6 years, 6 months ago
- 0 answers
Posted by Shivay Bishee 6 years, 6 months ago
- 1 answers
Posted by Jyoti Gadi 6 years, 6 months ago
- 1 answers
Yogita Ingle 6 years, 6 months ago
√3 = 1.73205....
√7 = 2.64575....
Therefore, 2 rational no. between √3 & √7 are;
- 2.01
- 2.1212121212121212....
Posted by Aniket Kumar Dubey Dubey 6 years, 6 months ago
- 2 answers
Yogita Ingle 6 years, 6 months ago
Total no. of outcomes; n(s) = 6= [1, 2, 3, 4, 5, 6]
No. Of favourable outcomes; n(e) = 2 = [3, 6]
P(getting a multiple of 3) = n(e)/ n(s)
= 2/6 = 1/3
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
