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Posted by Tushar Goswami 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
If your question is"Find the centre of a circle passing through the points (6,-6), (3,-7) and (3,3)." than the solution is-
Let A (6, –6), B (3, –7) and C (3, 3) be the points on the circle and the centre of the circle be P (h, k).
∴ PA = PB = PC (Radius of the circle)
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Posted by Anju Goel 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
Minimum value of an equation of the form a cosx+b sinx+c is
{tex}c-\sqrt{a^2+b^2}{/tex}
Here,c=8 a=3 b=4
=> minimum value= {tex}8-\sqrt{3^2+4^2}{/tex}
=> minimum value= 8 - 5= 3
Posted by Avni Gupta 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
Find the prime factorization of 150. 150 = 2 × 3 × 5 × 5.
Find the prime factorization of 210. 210 = 2 × 3 × 5 × 7.
LCM = 2 × 3 × 5 × 5 × 7.
LCM = 1050.
Posted by Subhjot Kaur 6 years, 2 months ago
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Posted by Hari Prasad [email protected] 6 years, 2 months ago
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Yogita Ingle 6 years, 2 months ago
As the numbers are divisible by both 2 and 5,so their last digit will always be 0. hence the series are 110,120,130,....,990
here a=110,d=10,an=990
an=a+(n-1)d
990=110+(n-1)10
(n-1)10=880
n-1=88
n=88+1
n=89.
thus the number of natural numbers between 102 and 998 which are divisible by both 2 and 5 is 89.
Read more on Brainly.in - https://brainly.in/question/12757639#readmore
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Aadya Singh ? 6 years, 2 months ago
Cosec^2A + sin^2A + 2cosecA×sinA = 4 [by squaring on both sides]
Cosec^2 + sin^2 + 2×1/sin ×sin = 4
Cosec^2 + Sin^2 + 2 = 4
Cosec^2 + Sin^2 = 2..
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M.......... Duhan 6 years, 2 months ago
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