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Ask QuestionPosted by ☺️ ? 6 years, 10 months ago
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Posted by Deep Sidhu 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
(x+3) (x-1) = x-1/3
x2 - x + 3x - 3 = x-1/3
x2 + 2x - 3 = x-1/3
3x2 + 6x- 9 = x-1
3x2+ 5 x - 8 = 0
3x2+ 8x - 3x - 8 = 0
x(3x + 8) - 1(3x + 8) = 0
(3x + 8) (x - 1 ) = 0
x = -8/3 , 1
So the roots are -8/3 and 1
Posted by Chitresh Kumar 6 years, 10 months ago
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Shabnam Star 6 years, 10 months ago
Posted by Anu Angel 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Let AO = H
CD = OB = 60m
A'B = AB = 60 + H
In Δ AOD,
⇒ 120 + H = 3H
⇒ 2 H = 120
⇒ H = 60 m
Thus, height of the cloud above the lake = AB + A'B = 60 + 60 = 120 m.
</form>Posted by Sandhya Sodhani 6 years, 10 months ago
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Posted by Ankit Yadav 6 years, 10 months ago
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Posted by Ankit Yadav 6 years, 10 months ago
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Posted by Mokesh Kanna 6 years, 10 months ago
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Posted by Siva Srinivasan 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Suppose level of water rises by h metres in the pond. Then, clearly, water risen in the pond forms a cuboidal of dimensions {tex}80 \mathrm { m } \times 80 \mathrm { m } \times \mathrm { hm }.{/tex}
{tex}\therefore \quad 80 \times 50 \times h{/tex}= Volume of water displaced by 500 persons
{tex}\Rightarrow \quad 80 \times 50 \times h = 500 \times 0.04{/tex}
{tex}\Rightarrow{/tex} 4000h = 20
{tex}\Rightarrow \quad h = \frac { 1 } { 200 } \mathrm { m } = 0.5 \mathrm { cm }{/tex}
Posted by Anurag Kadao 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let three consecutive numbers be x, (x + 1) and (x + 2)
Let x = 6q + r 0 {tex}\leq r < 6{/tex}
{tex}\therefore x = 6 q , 6 q + 1,6 q + 2,6 q + 3,6 q + 4,6 q + 5{/tex}
{tex}\text { product of } x ( x + 1 ) ( x + 2 ) = 6 q ( 6 q + 1 ) ( 6 q + 2 ){/tex}
if x = 6q then which is divisible by 6
{tex}\text { if } x = 6 q + 1{/tex}
{tex}= ( 6 q + 1 ) ( 6 q + 2 ) ( 6 q + 3 ){/tex}
{tex}= 2 ( 3 q + 1 ) .3 ( 2 q + 1 ) ( 6 q + 1 ){/tex}
{tex}= 6 ( 3 q + 1 ) \cdot ( 2 q + 1 ) ( 6 q + 1 ){/tex}
which is divisible by 6
if x = 6q + 2
{tex}= ( 6 q + 2 ) ( 6 q + 3 ) ( 6 q + 4 ){/tex}
{tex}= 3 ( 2 q + 1 ) .2 ( 3 q + 1 ) ( 6 q + 4 ){/tex}
{tex}= 6 ( 2 q + 1 ) \cdot ( 3 q + 1 ) ( 6 q + 1 ){/tex}
Which is divisible by 6
{tex}\text { if } x = 6 q + 3{/tex}
{tex}= ( 6 q + 3 ) ( 6 q + 4 ) ( 6 q + 5 ){/tex}
{tex}= 6 ( 2 q + 1 ) ( 3 q + 2 ) ( 6 q + 5 ){/tex}
which is divisible by 6
{tex}\text { if } x = 6 q + 4{/tex}
{tex}= ( 6 q + 4 ) ( 6 q + 5 ) ( 6 q + 6 ){/tex}
{tex}= 6 ( 6 q + 4 ) ( 6 q + 5 ) ( q + 1 ){/tex}
which is divisible by 6
if x = 6q + 5
{tex}= ( 6 q + 5 ) ( 6 q + 6 ) ( 6 q + 7 ){/tex}
{tex}= 6 ( 6 q + 5 ) ( q + 1 ) ( 6 q + 7 ){/tex}
which is divisible by 6
{tex}\therefore {/tex} the product of any three natural numbers is divisible by 6.
Posted by Khushi Agrawal 6 years, 10 months ago
- 5 answers
Gaurav Seth 6 years, 10 months ago
Q.In a quadrilateral ΔBCD, ∠A+ ∠D = 90°. Prove that AC2 + BD2 = AD2 + BC2.
Given Quadrilateral ΔBCD, in which ∠A+ ∠D = 90°
To prove AC2 + BD2 = AD2 + BC2
Construct Produce AB and CD to meet at E.
Posted by Khushi Agrawal 6 years, 10 months ago
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Jatt Putt Aryan???? 6 years, 10 months ago
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Gungun_ U Cn Wish Me On 3 Feb? 6 years, 10 months ago
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Gungun_ U Cn Wish Me On 3 Feb? 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Let a be the first term and d be the common difference of the AP. Then,
Pranav Songra 6 years, 10 months ago
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Posted by Pankaj Uraon 6 years, 10 months ago
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Yogita Ingle 6 years, 10 months ago
Let the first three terms of the AP be (a-d), a and (a+d). Then,
(a -d) + a + (a + d) = 48
⇒ 3a = 38
⇒ a = 16
Now,
(a - d) × a = 4(a + d) + 12 (Given)
⇒ (16 -d) × 16 = 4(16 + d) + 12
⇒16d + 4d = 256 - 76
⇒20d = 180
⇒ d= 9
When a= 16, d= 9
⇒ a - d = 16 - 9 = 5
a + d = 16 + 9 = 25
Hence, the first three terms of the AP are 7, 16, and 25.
Posted by Nikita Sharma 6 years, 10 months ago
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M A Santos 6 years, 10 months ago
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Dikshaa Chaudhary 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Since ABCD is a parallelogram,
AB = CD ---- i)
BC = AD ---- ii)
It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.
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