@krum said:hit and trials(2)/s(5*2) = s(2)/s(10) = 2s(1)/s(5*1) = 1/5 = .2
but how did u get that it will be max at n=2 and min at n=1?
@jain4444 said:N is a 3-digit number that is a perfect square. When the first digit is increased by 1, the second digit is increased by 2, the third digit is increased by 3, the result is still a perfect square. Determine N.
361?
3+1 , 6+2 , 1+3 = 484
@abhishek.2011 said:let faculty be in middle_ _ _ _ f _ _ _ _now in 8 blanks we arrange them as 8!now let the 2 boys b togetherall can be arranged in 2*7!so 8!-2*7!
u hav done half d part right ..d only thing u forgot is to add the no .of cases were boys sit beside d facullty in both sides,.
_ _ _ B f B _ _ _..
so in dis case d no .of ways r 6!*2=1040.
wich if we add 2 8!-2*7! we get d correct OA!!
_ _ _ B f B _ _ _..
so in dis case d no .of ways r 6!*2=1040.
wich if we add 2 8!-2*7! we get d correct OA!!
@jain4444 said:N is a 3-digit number that is a perfect square. When the first digit is increased by 1, the second digit is increased by 2, the third digit is increased by 3, the result is still a perfect square. Determine N.
squares end in
0,1,4,9,6 and 5
now this is increased by 3
so it can either be 1 or 6
as 1+3 = 4
and 6+3 = 9
tried such 3 digit squares....
got 361...
@jain4444 said:N is a 3-digit number that is a perfect square. When the first digit is increased by 1, the second digit is increased by 2, the third digit is increased by 3, the result is still a perfect square. Determine N.
361
used hit and trial as perfect squares always end in 1,4,5,6,9..so last digit +3 should b 1 of d following wich d no.of possibilities..
used hit and trial as perfect squares always end in 1,4,5,6,9..so last digit +3 should b 1 of d following wich d no.of possibilities..
@anytomdickandhary said:concepts applied1. cos formula (b^2 + c^2 - a^2 = 2b*c*cosA2. Area of the triangles within same parallels are in the ratio of their basesApply cos formulaIn triangle BAE: cosA = (4^2 + 4^2 - (2rt6)^2))/2*4*4 = 1/4now use the value of cosA derived above to find the length of AD.In triangle BAD: cosA = (4^2 + AD^2 - 8^2))/2*AD*4 =>AD = 8Hence, AE=ED = 4.Apply area for triangles in same parallels logicgiven BF = 4CFput area(CDF) = A => area(BEF)=4Aarea(CDF)+area(BEF) = 0.5*area(parallelogram)=>area(parallelogram) = 10ANow easy to see that area(ABE) = area(EDF) = (10A)/4hence area(DEFC)/area(Parallelogram) = 7/20ATDH.
sir i gt d value of AD by trial using appolonius..bt dis was a perfect way !!!
Mandeep has to create a password having 5 distinct characters using at least 2 digits (from 7 to 9) and at least 2 English vowels (from A, E, I, O and U). No character other than digits from 7 to 9 and vowels of English alphabets are allowed in that password. If the password starts with a digit, then it must end with an alphabet and if it starts with an alphabet, then it must end with a digit. Find the number of possible passwords that Mandeep can create. ?
@scrabbler said:528?432 cases when the boys are on opposite sides from the faculty member and 96 when they are on the same side...Edited - solved for 4 girls, I just realised Umm....31680?regardsscrabbler
Explain . :-)
@mohitjain said:Mandeep has to create a password having 5 distinct characters using at least 2 digits (from 7 to 9) and at least 2 English vowels (from A, E, I, O and U). No character other than digits from 7 to 9 and vowels of English alphabets are allowed in that password. If the password starts with a digit, then it must end with an alphabet and if it starts with an alphabet, then it must end with a digit. Find the number of possible passwords that Mandeep can create. ?
2880
3!(2C1*4C2 + 2C2*4C1 )*2*3C1*5C1
3!(2C1*4C2 + 2C2*4C1 )*2*3C1*5C1
@mohitjain said:Mandeep has to create a password having 5 distinct characters using at least 2 digits (from 7 to 9) and at least 2 English vowels (from A, E, I, O and U). No character other than digits from 7 to 9 and vowels of English alphabets are allowed in that password. If the password starts with a digit, then it must end with an alphabet and if it starts with an alphabet, then it must end with a digit. Find the number of possible passwords that Mandeep can create. ?
There are only 2 possible cases
1>Either it starts with a digit
3*5*(4C1*3!+2C1*4C2*3!)
2>When it starts with an alphabet
3*5*(4C1*3!+2C1*4C2*3!)
So it should be 2*3*5(4C1*3!+2C1*4C2*3!)
3*5*(4C1*3!+2C1*4C2*3!)
So it should be 2*3*5(4C1*3!+2C1*4C2*3!)
@joyjitpal said:28803!(2C1*4C2 + 2C2*4C1 )*2*3C1*5C1
@sam05 said:There are only 2 possible cases1>Either it starts with a digit3*5*(4C1*3!+2C1*4C2*3!)2>When it starts with an alphabet3*5*(4C1*3!+2C1*4C2*3!)So it should be 2*3*5(4C1*3!+2C1*4C2*3!)
its 2880!!!
[x] is the greatest integer function. It denotes the greatest integer value of x that is not greater than x.
What is the area enclosed by y = 0, x = 4 and y = [x] + 5x?
What is the area enclosed by y = 0, x = 4 and y = [x] + 5x?
1) 26 square units
2) 6 square units
3) 46 square units
4) None of these
2) 6 square units
3) 46 square units
4) None of these
@shinoda said:[x] is the greatest integer function. It denotes the greatest integer value of x that is not greater than x.What is the area enclosed by y = 0, x = 4 and y = [x] + 5x?1) 26 square units2) 6 square units3) 46 square units4) None of these
46???
Done using graph
Done using graph
@shinoda said:[x] is the greatest integer function. It denotes the greatest integer value of x that is not greater than x.What is the area enclosed by y = 0, x = 4 and y = [x] + 5x?1) 26 square units2) 6 square units3) 46 square units4) None of these
You will need to take values of x...as in cases...
From 0 to 1..
1 to 2..
2 to 3..
3 to 4...
@shinoda said:can u help realise the graph of [x]+5x ?
Take cases...it should be a straight line...
@shinoda said:can u help realise the graph of [x]+5x ?
For eg...
When x from 0 to 1..
It will be 5x..
From 1 to 2
1 + 5x...
@shinoda said:can u help realise the graph of [x]+5x ?
For eg...
When x from 0 to 1..
It will be 5x..
From 1 to 2
1 + 5x...
@shinoda said:can u help realise the graph of [x]+5x ?
hey sorry for the late reply..actually what u can do is take each interval separately and integrate..
Hence 0 to 1 me the function would be 0+5x where x from 0 to 1..
so now integrate this..
similarly for 1 to 2 its 1+5x and interval 1 to 2...so integrate
bakika b ese nikalke just add the values
Hence 0 to 1 me the function would be 0+5x where x from 0 to 1..
so now integrate this..
similarly for 1 to 2 its 1+5x and interval 1 to 2...so integrate
bakika b ese nikalke just add the values