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Sia ? 5 years, 5 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 5 years, 5 months ago
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Posted by Diya Garg 5 years, 5 months ago
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Sia ? 5 years, 5 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 5 years, 5 months ago
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Sia ? 5 years, 5 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 5 years, 5 months ago
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Posted by Krish Oberoi 5 years, 5 months ago
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Sia ? 5 years, 5 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 5 years, 5 months ago
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Yogita Ingle 5 years, 5 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 ? 5 years, 5 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 5 years, 5 months ago
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Sia ? 5 years, 5 months ago
6x2 + 17x + 5
= 6x2 + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1)(2x + 5)
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Yogita Ingle 5 years, 5 months ago
(x + a) (x + b) = x2 + (a+b)x + ab
(X+4)(x+10) = x2 + (10+4)x + 10(4)
= x2 + 14x + 40
2Thank You