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Ask QuestionPosted by Shriyanshi Gupta 6 years, 10 months ago
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Rishab Jain 6 years, 10 months ago
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Honey ??? 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
2x - 1= 0
x = 1/2
So, by putting the value of x = 1/2
4x3 - 12x2 + 14x - 3
= 4 (1/2 )3 - 12 (1/2 )2 + 14 (1/2 ) - 3
= 4 (1/8) - 12 (1/4) + 14/2 - 3
= 4/8 - 12/4 + 14/2 - 3
= 1/2 - 3 + 7 - 3
= 1/2 + 1
= 1+2 / 2
= 3/2 - Ans
So the remainder is 3/2 .
Posted by Avaneesh Sharma Mdjssv 6 years, 10 months ago
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Posted by Aayushi . 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
As, volume of hemisphere = 2425{tex}\frac { 1 } { 2 } c m ^ { 3 }{/tex}
{tex}\Rightarrow \frac { 2 } { 3 } \pi r ^ { 3 } = 2425 \frac { 1 } { 2 }{/tex}
{tex}\Rightarrow \frac { 2 } { 3 } \times \frac { 22 } { 7 } r ^ { 3 } = \frac { 4851 } { 2 }{/tex}
{tex}\Rightarrow r ^ { 3 } = \frac { 4851 \times 3 \times 7 } { 2 \times 2 \times 22 }{/tex}
{tex}\Rightarrow r ^ { 3 } = \frac { 441 \times 3 \times 7 } { 2 \times 2 \times 2 }{/tex}
{tex}\Rightarrow r ^ { 3 } = \frac { 21 ^ { 3 } } { 2 ^ { 3 } }{/tex}
{tex}\Rightarrow r = \frac { 21 } { 2 } c m{/tex}
So, the curved surface area of the hemisphere ={tex}2 \pi r ^ { 2 }{/tex}
{tex}= 2 \times \frac { 22 } { 7 } \times \frac { 21 } { 2 } \times \frac { 21 } { 2 } = 693 \mathrm { cm } ^ { 2 }{/tex}
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Gaurav Seth 6 years, 10 months ago
Correct question: The points A(0,3), B(-2, a) and C(-1,4) are the vertices of a right angled triangle at A. Find the value of a.
<hr />Solution:
Length of AB =√[(-2-0)²+(a-3)²]=√4+(a-3)²
Length of BC=√1²+(4-a)²=√1+4-a²
Length of CA=√1²+1²=√2
BC²=AC²+AB²
1+(4-a)²=2+4+(a-3)²
1+16+a²-8a=6+a²+9-6a
17+a²-8a=15+a²-6a
2+a²-a²=-6a+8a
2a=2
a=1
Puja Sahoo? 6 years, 10 months ago
Posted by Hariom Tiwari 6 years, 10 months ago
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Yogita Ingle 6 years, 10 months ago
Let the points (5, - 2), (6, 4), and (7, - 2) are representing the vertices A, B, and C of the given triangle respectively.
AB = {tex}\;\sqrt{{(5-6)}^2+\;(-2-4)^2}\;=\;\sqrt{(-1)^2\;+\;(-6)^2}\;=\;\sqrt{1\;+36\;}\;=\sqrt{37}{/tex}
BC = {tex}\;\sqrt{{(6-7)}^2+\;(4\;-\;(-2))^2}\;=\;\sqrt{(-1)^2\;+\;(6)^2}\;=\;\sqrt{1\;+36\;}\;=\sqrt{37}{/tex}
CA = {tex}\;\sqrt{{(5-7)}^2+\;(-2)\;-\;(-2))^2}\;=\;\sqrt{(-2)^2\;+\;(0)^2}\;=\;\sqrt{4\;}\;=2{/tex}
Therefore, AB = BC
As two sides are equal in length, therefore, ABC is an isosceles triangle.
Puja Sahoo? 6 years, 10 months ago
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Rishab Jain 6 years, 10 months ago
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