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Ask QuestionPosted by Rekha Vyas 6 years, 9 months ago
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Posted by Yash Sharma ?? 6 years, 9 months ago
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Yogita Ingle 6 years, 9 months ago
First term of A.P. = 9 , common difference = 8
if sum of n terms 636, Sn = (n/2) [ 2×9 + (n-1)8 ] = 636
Sn = n [ 10+ 8n] = 1272
8n2 +10n -1272 = 0
4n2 +5n -636 = 0
By factorising LHS,
(4n+53)(n-12) = 0
hence n = 12
Posted by Faiz Saikh 6 years, 9 months ago
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कबीर गुर्जर अभिषेक का दसवाँ बाप 6 years, 9 months ago
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#Aditi~ Angel???? 6 years, 9 months ago
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Posted by Ship.. ??? 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Given, Three cubes of a metal whose edges are in the ratio 3 : 4 : 5{tex}{/tex}.So,we can take the edges of these cubes as 3x, 4x and 5x.
Volume of the cones are, (3x){tex}^3{/tex}, (4x){tex}^3{/tex}, (5x){tex}^3{/tex} i.e 27x{tex}^3{/tex}, 64x{tex}^3{/tex}, 125x{tex}^3{/tex}.
We know that,
the length of the diagonal of a single cube of side {tex}a = a \sqrt { 3 }{/tex}
The length of the diagonal of a cube is given to be {tex}12 \sqrt { 3 }.{/tex}
So, the edge of a single cube is 12 cm.
Volume of the single cube = (12)3 = 1728 cm3
Since the three cubes are melted and converted into a single cube,
Sum of the volumes of the three cubes = Volume of a single cube
{tex}\Rightarrow{/tex} 27x3 + 64x3 + 125x3 {tex}=1728{/tex}
{tex}\Rightarrow{/tex} 216x3 {tex}= 1728{/tex}
{tex}\Rightarrow{/tex} x3 = 8
{tex}\Rightarrow{/tex} {tex}x = 2 \ cm{/tex}
So, the edges of the cubes are 6 cm, 8 cm and 10 cm.
Posted by Nisha Sandhu (Super Cool) 6 years, 10 months ago
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Posted by Suman Chimpa 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We have,
xsin3{tex}\theta{/tex} + ycos3{tex}\theta{/tex} = sin{tex}\theta{/tex} cos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} (xsin{tex}\theta{/tex}) sin2{tex}\theta{/tex} + (ycos{tex}\theta{/tex}) cos2{tex}\theta{/tex} = sin{tex}\theta{/tex} cos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} x sin{tex}\theta{/tex} (sin2{tex}\theta{/tex}) + (x sin{tex}\theta{/tex}) cos2{tex}\theta{/tex} = sin{tex}\theta{/tex} cos{tex}\theta{/tex} [{tex}\because{/tex} x sin{tex}\theta{/tex} = y cos{tex}\theta{/tex}]
{tex}\Rightarrow{/tex} x sin{tex}\theta{/tex} (sin2{tex}\theta{/tex} + cos2{tex}\theta{/tex}) = sin{tex}\theta{/tex} cos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} x sin{tex}\theta{/tex} = sin{tex}\theta{/tex} cos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} x = cos{tex}\theta{/tex}
Now, xsin {tex}\theta{/tex} = ycos{tex}\theta{/tex}
{tex}\Rightarrow{/tex} cos{tex}\theta{/tex} sin{tex}\theta{/tex} = y cos{tex}\theta{/tex} [{tex}\because{/tex} x = cos{tex}\theta{/tex}]
{tex}\Rightarrow{/tex} y = sin{tex}\theta{/tex}
Hence, x2 + y2 = cos2{tex}\theta{/tex} + sin2{tex}\theta{/tex} = 1
Posted by Shehnaz .... 6 years, 10 months ago
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