Algebra, Quadratic Equations, Inequalities, Functions

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The number of real roots of the equation |1 - |x|| - (1.01)(1.01x) = 0 is/are
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HOW TO SOLVE??


 Ashok went on a tour. He visited a total of 8 cities. In each city he spent र2 less than half the amount he had with him. He spent र100 in the last city he visited. Find the amount he had initially (in र). 



ANSWER IS 25604

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Ashok went on a tour. He visited a total of 8 cities. In each city he spent र2 less than half the amount he had with him. He spent र100 in the last city he visited. Find the amount he had initially (in र).

12804
25604
24580
24804
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This is an extremely easy question but I am missing sth while solving it algebraically. Pls help.

|y|+|y+2|=2................Solve for y

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kimig
@kimig  ·  166 karma

plzz give the exact solution with values and the correct answer too

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priyanjit
@priyanjit  ·  322 karma

y = [-2,0] . First consider for positive values, hence : y+y+2 =2 , so y = 0 and then consider for negative values, -y -y-2 =2 , so y= -2, so that is the range. I hope this is useful.

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any link or page on reflection of points/graphs specially for quadratic or cubic eq. ??

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A=log10(1+2+3+.....................n)+log10^2. where n is a natural number .find the number of possible value of n for which but range is 1

(1)28

(2)31
(3)29
(4)38

2 comments
foadbear
@foadbear  ·  22,638 karma

confused , what does it means, "for which but range is 1" 

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priyanjit
@priyanjit  ·  322 karma

is it 10 multiplied by (1+2+3..n) or it is base 10.

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How many 4-digit postive integral number are their base?If the number of such number is converted to the same base?

(1)2058
(2)5666
(3)6000
(4)NOTA

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karan77sethi
@karan77sethi  ·  30 karma

if any can do pl

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IF x=12, then the value of x^5-13x^4+13x^3-13x^2+13x-1?

(1)0

(2)1

(3)2

(4)3

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foadbear
@foadbear  ·  22,638 karma

just factorized it and took out x 

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Find the number of real solution of a equation is

x=(x-1/2)^(1/2)+(1-1/2)^(1/2)

(1)0

(2)1

(3)2

(4)3

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rashmi84
@rashmi84  ·  0 karma

1

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AV91
@AV91  ·  0 karma

second term is non zero +ve while first term can not be negative because square root can't be -ve hence  no solution.

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The reflection of the graph y=(x+3)(x+4) in the point (2,2) is


-x^2+15x+52
-x^2+15x-52
x^2+15x-52
x^2+15x+52
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The reflection of the graph y=(x+3)(x+4) in the point (2,2) is


-x^2+15x+52
-x^2+15x-52
x^2+15x-52
x^2+15x+52
2 comments5 answers
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