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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
(3) Determine the value of x + y if3) Determine the value of x= vir 3 a 41-17 73=13][2x+y5x -74x4x(7 70-137Ly x+6s |
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Answer» The answers are x=2 and y= 3 |
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| 2. |
((-7)^3*(1/(-7))^(-9))/(-7)^10 |
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| 3. |
2. Additive inverse of -3-4iis(a) - 3+4i.(b) +3+41.(c) 3141.(d) +3-41 |
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Answer» option B because addititive inverse means answer becoming 0 |
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| 4. |
1- 0LLLL LL U.(200133 If P is the solution set of - 3x + 4 < 2x - 3, XE N, and4x – 5 < 12, xe W, findis the solution set ofon the |
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| 5. |
(2011)32 If x E I, A is the solution set of 2(x - 1) < 3x - 1 and B is the solution set of4x - 3 3 8 + x, find A B. |
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| 6. |
9. Write the solution set of x + |
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Answer» Modulus question...... 👍👍👍👍👍👍❓❓❓ | x + 1/x|>0; x^2+1/x>2; x^2+1>2x,; x^2-2x-1=0; x^2-x-x-1=0; x( x-1)+1( x-1); ( x-1)( x+1) ; x=-1 & 1 |
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| 7. |
28.Find the solution set of the equation |
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| 8. |
Ifx + = 3, calculate xPand x* ++ |
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Answer» check outmy solution |
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| 9. |
Find the solution of the inequalityHence, graph the solution set on the number line. |
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| 10. |
Write all the proper subsets of the given setA = (x:x is a letter of the word "GOD"} |
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| 11. |
(a)25value of 1 + 4 × 34 × 32 , 4 × 33is |
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Answer» 1+1/12+1/36+1/108 1.12 |
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| 12. |
2Write the solution set of the inequation> 0. |
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| 13. |
32 ÷ 4 + 6 × 12 |
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Answer» 32 ÷ 4 + 6 × 12 = 8 + 72 = 80 Thank you |
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| 14. |
(32)^{-4 / 5} |
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Answer» 32 = 2 power 5so32 power ( -4/5)2^ (-4×5)/52^(-4)1/2^41/16 this is not the answer |
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| 15. |
рдиреАo1 | S~ |~ |
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Answer» please like my answer if you find it useful |
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| 16. |
s‘—\"“:F< o1 |
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| 17. |
\left(\frac{4}{3}\right)^{2 y+9} \times\left(\frac{4}{3}\right)^{y+1}=\left(\frac{4}{3}\right)^{1} |
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| 18. |
( i s - a ) - \frac { 4 } { 3 } , b ) \frac { 4 } { 3 } , c ) \frac { 4 } { 5 } , d ) - \frac { 4 } { 5 } |
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Answer» tant=3so cost=1/root(9+1)=1/root(10)cos2t=2cos^2t-1=2(1/root(10))^2-1 =2(1/10)-1=1/5-1=-4/5 |
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| 19. |
| \frac { - 4 } { 3 } + | \frac { 5 } { 8 } | \text { and } | \frac { - 4 } { 3 } | + | \frac { 5 } { 8 } | |
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| 20. |
2 \frac { 1 } { 3 } - \frac { 3 } { 4 } ( 6 ) 7 \frac { 1 } { 6 } \div \frac { 5 } { 6 } |
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| 21. |
Eydmple 29 Ifx = 2 + 18, find the values of:x2(ii) x2- |
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| 22. |
\frac { 3 \frac { 1 } { 4 } - 2 \frac { 1 } { 8 } } { 6 \frac { 1 } { 4 } + 3 \frac { 1 } { 6 } } |
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| 23. |
32. If A(1, 2, 3, 4, 5), then the number of proper subsets of A is(A) 120(B) 30(C) 31(D) 32 |
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| 24. |
- /% 7+l02âl+\l4+....+84 |
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Answer» Clearly it is an A.P. Here, a=7 ,d=7/2 , an=84 , Sn=? We know that,S = n/2 (a+l)S = 23/2 (7+84)S = (23/2) * 91S = 2093/2= 1046.5 So, the sum of 7+21/2+14+.............+84 is 1046.5 hit like if you find it useful |
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| 25. |
\frac { 5 \sin ^ { 2 } 30 ^ { \circ } + \cos ^ { 2 } 45 ^ { \circ } - 4 \tan ^ { 2 } 30 ^ { \circ } } { 2 \sin 30 ^ { \circ } \cos 30 ^ { \circ } + \tan 45 ^ { \circ } } |
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| 26. |
P (L4 ) =P+ ) |
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Answer» dy/dx=x2+y2/x2+xy -(1) Thus, dy/dx is a homogeneous function. Put y=vx in the eq. dy/dx=v+xdv/dx Put in eqn.(1), v+xdv/dx=x2+y2/x2+xy on solving, we get, 1+v/1-vdv=1/xdx Integrating both sides we get, -2log|v-1|-v=log|x|+c Put v=y/x, -2log|(y/x)-1|-y/x=log|x|+c, |
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| 27. |
1 , 7 goats cost? 630. Find the cost of 1 goat |
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Answer» Cost of 7 goats = ₹ 630Cost of 1 goat = 630/7 = ₹ 90. Please hit the like button if this helped you |
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| 28. |
\frac{5 \sin ^{2} 30^{\circ}+\cos ^{2} 45^{\circ}-4 \tan ^{2} 30^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\tan 45^{\circ}} |
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Answer» PLEASE LIKE THE SOLUTION PLEASE LIKE THE SOLUTION |
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| 29. |
AMPLE-7. Determine the value of.-\frac{5 \sin ^{2} 30^{\circ}+\cos ^{2} 45^{\circ}-4 \tan ^{2} 30^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\tan 45^{\circ}} |
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Answer» sin30=1/2 cos45=1/√2 tan30=1/√3 cos30=√3/2hence the given expression5/4+1/2-4/3/1*√3/2+114/8-4/3/√3+2/242-32/24/√3+2/210/12(√3+2) |
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| 30. |
(2) Height and weight proportion (BMI): |
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Answer» Body Mass Index (BMI) is a person's weight in kilograms divided by the square of height in meters. A high BMI can be an indicator of high body fatness. The body mass index or Quetelet index is a value derived from the mass and height of an individual. The BMI is defined as the body mass divided by the square of the body height, and is universally expressed in units of kg/m², resulting from mass in kilograms and height in metres. Please hit the like button if this helped you out. |
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| 31. |
20% of dolls produced in a factory weredetective, 25% of the remaining were damaged.If 4800 dolls were left over, what was the originalnumber of dolls? |
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Answer» thanks right answer hai |
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| 32. |
20% of dolls produced in a factory wedefective, 25% of the remaining were damagIf 4800 dolls were left over, what was the orignumber of dolls? |
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| 33. |
18 dolls cost Rs 630, how many dolls can be bought for Rs 455 |
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| 34. |
( \frac .32 243 ) ^ \frac 4 3 |
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| 35. |
(ii) Lt I+xx2+10 |
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Answer» please like my answer if you find it useful |
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| 36. |
1 *23.If-+x2 +1, findx2 - 1 |
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| 37. |
0Dy uIs = सा + ‘_(L.___v'.”) (o1) |
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Answer» please like if you find it useful |
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| 38. |
\frac { 5 \cos ^ { 2 } 60 ^ { \circ } + 4 \sec ^ { 2 } 30 ^ { \circ } - \tan ^ { 2 } 45 ^ { \circ } } { \sin ^ { 2 } 30 ^ { \circ } + \cos ^ { 2 } 30 ^ { \circ } }° |
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| 39. |
. If 18 dolls cost Rs.630, how many dolls can be bought for Rs.455? |
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| 40. |
5 hours?6. If 18 dolls cost 630, how many dolls can be bought for 455? |
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| 41. |
19, D and E are points on the sides AB and AC respectively of a ΔABC such that, DEJBC anddivides AABC into two parts equal in area. Prove that, AB 2 |
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Answer» thanks |
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| 42. |
(vi) x*+2(cosec )Xyt y2=0 |
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Answer» Thanks bhai |
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| 43. |
the cost oCDEB)- ar (AABC15 In the figure of AABC, D divides AC in the ratio 4:3. If DE BC, then find ar (BCDE) ar (AABC)or that D and H lie |
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Answer» thank you so much 😊😊😊 |
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| 44. |
= x 'I +xĘ =(x)dI SITANOTANTE |
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Answer» 1 is the correct answer of the given question p(-1/3)=0 0 is the best answer |
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| 45. |
\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}} |
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| 46. |
शिव. किशा bys L4 3ds-2 .Jc : |
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Answer» Rationalize[(4+3√5)^2]/(16-45)(16+45+25√5)/(-29)(61+25√5)/(-29) a=-61/29b=-25/29 |
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| 47. |
ŕ¤ŕ¤°:16. In the figure, findL1+ L2+ L3+ L4+ LS. |
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Answer» can you understand now? first write this If not send Then I will send again first this then that |
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| 48. |
âŹ7 = AS'0 + 01 = Ag0 + XT0 |
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Answer» 0.2x+0.3y=1.30.4x+0.5y=2.3 Multiply equation 1 with 2 0.4x+0.6y=2.60.4x+0.5y=2.3 Subtract0.1y=o.3y=3 x=1.3-0.9/0.2=.4/.2=2 x=2, y=2 thanks sir |
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| 49. |
Difference the following fuction\frac{-1}{6 x^{6}}-\frac{1}{5 x^{5}}-\frac{1}{4 x^{4}} |
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Answer» thank you for answering |
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| 50. |
\left(\frac{32}{243}\right)^{-4 / 5}=? |
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Answer» नाही समजलं परत एकदा |
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