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sum of 5 consecutive whole numbers, x + (x+1) + (x+2) + (x+3) + (x+4) =2005 5x + 10 = 2005 find x from here and write each term and then add their digits simply.
Already answered this question. Plz do check again! Write, {y^2 - (3/24)^2} {y^2-(1/24)^2} as => {y^2 - (1/24)^2 – 8/(24)^2} {y^2-(1/24)^2} = 5/864 let {y^2 - (1/24)^2} = xthen,(x -8/(24)^2) x =...
Write, {y^2 - (3/24)^2} {y^2-(1/24)^2} as => {y^2 - (1/24)^2 – 8/(24)^2} {y^2-(1/24)^2} = 5/864 let {y^2 - (1/24)^2} = xthen,(x -8/(24)^2) x = 5/864Now solve this quadratic in x and then equate it to ...
The minimum value of the equation is given by -D/4a. D = A^2-4AB, so -D = 4AB – A^2 4a = 4, so -D/a=4a = 4AB-A^2/4. Since A and B are the roots, 4AB-A^2 = 0 So minimum value of equation is equal to 0
Can you please tell the rough figure for this question with proper labelling of sides as i am not able to get it
sol
Several ways to approach this: Write where So since Hence So either z is purely imaginary or z=0. (z=0 is possible when z 1 =-z 2 which is permitted with the restrictions mentioned in the problem)
From AM-GM inequality with equality when b=c. Repeat this for the other two terms and add and you will get RHS with equality when a=b=c
Total students when they meet at the same hour. 60 + 40 = 100. In this 100 students the common would be: 60 – 40 = 20. the AUB = A + B – AintersectionB = > 60 + 40 – 20 = > 80
Given, x + (x+1) + (x+2) + (x+3) + (x+4) =2005 5x + 10 = 2005 5x = 1995 x = 399 Numbers are therefore, 399, 400, 401, 402, 403 Now u can sum their digits.
First find the angle b/w the eqns. by using the formulae: tan(thetha) = |m1 – m2|/|1 + m1*m2| m1 and m2 are the respective slopes of the two eqns. then take the obtuse angle. then half of that angle...
Hi, Please find the solution 2 2 +5 2 +8 2 +......15 terms =2 2 +5 2 +8 2 +.....+44 2 Now solve this
factorial n is already defined as the product of all positive numbers less than or qual to n. This is the definition n factorial. and 0 factorial is 1.
Conditions: 25 – x^2 should be greater and equal to zero solve for x (with wavy curve and find the interval). and sin(cosx) should be greater than equal to zero. or cosx greater than equal to...
-1)=57 x – 65(2 2x+3 Write: 2 as (2 x )^2.8 – 65(2 x -1)=57 => and assume 2^x = t => 8t^2 – 65t = 57 – 65 => 8t^2 – 65t = -9 solve the quadratic.
Sum of p terms, Sp = p/2 [2a + (p-1)d] Sum of q terms, Sq = q/2[2a + (q-1)d] And given Sp/Sq = p^2/q^2 do it and sort out further u would get somthing something relation in pth and qth terms relation.
are perpendicular i.e. for some angle Then Then its immediately seen that and w 1 and w 2 are also perpendicular. Which means that
If the point (H,K ) lies on the line 2x+3y=6 , the smallest value of H square+ k square would be as follows: put (H,K) in the given eq. 2h + 3k = 6 or k = (6 – 2h)/3 Now, let M = h^2 + k^2 put value...
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