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Gaurav Seth 5 years, 4 months ago
The first 3-digit number which is divisible by 7 is 105
The last 3-digit number which is divisible by 7 is 994
The list of 3-digit numbers divisible by 7 are
105, 112, 119,…..994 which forms an A.P
Now using the A. P Formula,
T(n) = a + (n – 1)d
Where
a = 105
d = 7
T(n) = 994
994 = 105 + (n – 1)7
889 = 7n – 7
7n = 896
n = 128
∴ There are 128 3-digits number which are divisible by 7.
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Gaurav Seth 5 years, 4 months ago
Given : 15x² - 41x + 14 = 0
15x2 - 35x - 6x - 14 = 0
5x (3x-7) + 2 (3x-7) = 0
(5x-2) = 0 (3x-7) = 0
x = 2/ 5
or
x = 7/3
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Gaurav Seth 5 years, 4 months ago
QUESTION: Solve; 1/a+b+x=1/a+1/b+1/x
Answer:
1/(a+b+x)=1/a + 1/b + 1/x
1/(a+b+x) -1/x = 1/a + 1/b
x-(a+b+x)/x(a+b+x) = a+b/ ab
-1(a+b) /x(a+b+x) = a+b/ ab
-1(a+b)/(a+b) = x(a+b+x)/ab
-1=x(a+b+x) / ab
-ab= (a+b)x + x2
so, x2 + (a+b)x +ab = 0
now, A = 1, B=(a+b) C= ab..
so, D = B2 - 4AC
=(a+b)2 - 4ab
= (a-b)2
so, x = (-B + root D)/2A and (-B-root D)/2A
= -a-b+a-b /2*1 and -a-b-a+b /2*1
= -2b/1 and -2a /1
so,x = -2b or -2a .
Posted by Tanmay Bagade 5 years, 4 months ago
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Yogita Ingle 5 years, 4 months ago
x²-4=0
x2 - (2)2 = 0
(x - 2) (x + 2) =0
x - 2 = 0 and x + 2 = 0
x = 2 and x = -2
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