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| 16351. |
Which practise book will good for maths board exam? |
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Answer» Rd Sharma and RS Aggarwal and examidea Mtg cbse champion Reply kr dio Prachi I think evergreen or you can choose whatever Prachi |
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| 16352. |
Hlo..anyone online?? |
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Answer» Say ur time Ya When u will be active??? Mm Ok Its voldemort Ur profile pic??? Ok.no prblm Actually iam not active Ok. Ak fabz My user name is aishreet.27 and name is arshdivya...in profile.pic.there is black and white pic.of earings. Oh wow..can u tell me ur id..if u dont mind Tell ur name Yeah Going good...and urs How is +1 R u on insta. Actually i m on Instagram now..thats y m asking Fine???.alive Fine How r u? How r u Same to uu Happy frndship day Hey Hi snehil Oh hi alkesh..yes its me Is this aishreet Yes |
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| 16353. |
Find the HCF of the following by using long division method |
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| 16354. |
Ex 6.3. Question---14 |
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| 16355. |
What is sin(sin(sin(sin...........))) |
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Answer» ?? You\'n do your Time- Pass at some other website Too...... kyonki yahaan toh sirf exam mein log ate hain naheen toh bahot boring website Hai..... Actually my ques is , how to do timepass and thats what i am doing ? |
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| 16356. |
X^0 = 1 Why |
| Answer» X^0= x^(1-1) =(X^1)/(x^1) =1 | |
| 16357. |
Ncert ex- 6.5, question no. 2,8,11 |
| Answer» You can check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 16358. |
16 n can end with 0 for n belongs to N |
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Answer» It cannot end with digit zero because in 16 n (2×2×2×2) n is there and for ending with digit zero the number must contain 2 and 5 in its prime factors and 16 only contain 2 but not 5 that is why it cannot end with 0 16ⁿ = (2×2×2×2)ⁿPrime factorisation of 16ⁿ does not contain the prime factor 5. Therefore, 16ⁿ cannot end with digit 0 for any natural number n |
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| 16359. |
Prove that √1+cos/1-cos=1+cos/sin |
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Answer» Multiply both numerater and denominator by 1+cosYou will get √(1+cos)^2/√1-cos^21-cos^2 = sin^2√(1+ cos)^2/√sin^21+cos/sin √1+cos 1. --------------- √1-cos 2. √1+cos x1+cos -------------------------- √1-cos x 1+cos 3. √(1+cos)*2 ------------------ √ 1-cos*24. √1+cos ---------- √ Sin*25. 1+cos ------- Sin |
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| 16360. |
Hii...? |
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Answer» Hlo bestie Hlo |
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| 16361. |
X+y/2 +y-1/3=9X-1/3 + y+1/2 =8 |
| Answer» The given equations may be written as{tex} \\frac { x + 1 } { 2 } + \\frac { y - 1 } { 3 } = 8 {/tex}⇒ 3(x + 1) + 2 (y - 1) = 48⇒ 3x + 3 + 2y - 2 = 48⇒ 3x + 2y + 1= 48⇒ 3x+2y = 47 ... (i){tex}\\frac { x - 1 } { 3 } + \\frac { y + 1 } { 2 } = 9{/tex}⇒ 2(x -1 ) + 3(y +1) = 54⇒ 2x - 2 + 3y + 3 = 54⇒ 2x + 3y + 1 = 54⇒ 2x + 3y = 53. ... (ii)Multiplying (i) by 2 and (ii) by 3 and subtracting, we get(4\xa0- 9)y = 94-159{tex} \\Rightarrow - 5 y = - 65 {/tex}{tex}\\Rightarrow y = \\frac { - 65 } { - 5 } {/tex}{tex}\\Rightarrow y = 13{/tex}Putting y = 13 in (i), we get3x\xa0+ (2 {tex} \\times{/tex}\xa013) = 47{tex} \\Rightarrow{/tex} 3x + 26 = 47{tex} \\Rightarrow{/tex} 3x = (47 - 26){tex} \\Rightarrow{/tex}3x = 21{tex} \\Rightarrow \\quad x = \\frac { 21 } { 3 } = 7{/tex}So, x = 7Hence, x = 7 and y = 13 | |
| 16362. |
In a lottery there are 8 prizes and 16 blanks.What is the probability of getting a price? |
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Answer» 8+16= 248/24== 1/3 Go ahead and use youtube 50% |
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| 16363. |
If perimeter and area of a circle is equal and find radius |
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Answer» 2 Radius is 2 2 |
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| 16364. |
X2+5x-(a2+a-6)=0 |
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| 16365. |
Two opposite vertices of a square (-1,2) and (3,2). Find the coordinate of remaining vertices. |
| Answer» Let ABCD be a square and B (x, y) be the unknown vertex.AB = BC{tex} \\Rightarrow {/tex} AB2 = BC2\xa0{tex} \\Rightarrow {/tex}\xa0(x + 1)2 + (y - 2)2 = (x - 3)2 + (y - 2)2{tex} \\Rightarrow {/tex} x2 + 1 + 2x + y2 + 4 - 4x = x2 - 6x + 9 + y2 + 4 - 4x{tex} \\Rightarrow {/tex}\xa02x + 1 = - 6x + 9{tex} \\Rightarrow {/tex}\xa08x = 8{tex} \\Rightarrow {/tex}\xa0x = 1 ........ (i)In {tex}\\triangle{/tex}ABC, AB2 + BC2 = AC2{tex} \\Rightarrow {/tex}\xa0(x + 1)2 + (y - 2)2 + (x - 3)2 + (y - 2)2 = (3 + 1)2 + (2 - 2)2{tex} \\Rightarrow {/tex}x2 + 1 + 2x + y2 - 4x + 4 + x2 + 9 - 6x + y2 + 4 - 2y = 16 + 0{tex} \\Rightarrow {/tex}\xa02x2 + 2y2 + 2x - 4y - 6x - 4y + 1 + 4 + 9 + 4 = 16{tex} \\Rightarrow {/tex}\xa02x2 + 2y2 - 4x - 8y + 2 = 0{tex} \\Rightarrow {/tex}\xa0x2 + y2 - 2x - 4y + 1 = 0 ..... (ii)Putting the value of x in eq. (ii),1 + y2 - 2 - 4y + 1 = 0\xa0{tex} \\Rightarrow {/tex}\xa0y2 - 4y = 0\xa0{tex} \\Rightarrow {/tex}\xa0y(y - 4) = 0{tex} \\Rightarrow {/tex}\xa0y = 0 or 4Hence the other vertices are (1, 0) and (1, 4). | |
| 16366. |
Sin18°/cos72° evaluate this. |
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Answer» Sin 18/cos72 cos(90-72)/cos72Cos72/sin72=1Tan26/tan(90-74)=1Cos48-sin42=sin(90-42)-sin42=0Cosec31-Cosec(90-69) Cose31-Cosec31=0 Sin theata = cos (90 - theata) So, Sin 18° =cos (90°- 18°) =cos 72° Cos 72°/cos72° =1 |
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| 16367. |
Given seco=13/12, calculate all trigonometrc ratios. |
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Answer» Consider a triangle ABC in which {tex}\\angle A = \\theta {/tex}\xa0and {tex}\\angle B = {90^o}{/tex}Let AB = 12k and AC = 13kThen, using Pythagoras theorem,\xa0{tex}BC=\\sqrt { ( \\mathrm { AC } ) ^ { 2 } - ( \\mathrm { AB } ) ^ { 2 } } = \\sqrt { ( 13 k ) ^ { 2 } - ( 12 k ) ^ { 2 } }{/tex}{tex}= \\sqrt { 169 k ^ { 2 } - 144 k ^ { 2 } } = \\sqrt { 25 k ^ { 2 } } = 5 k{/tex}{tex}\\therefore {/tex}\xa0{tex}\\sin \\theta = \\frac { B C } { A C } = \\frac { 5 k } { 13 k } = \\frac { 5 } { 13 }{/tex}{tex}\\cos \\theta = \\frac { A B } { A C } = \\frac { 12 k } { 13 k } = \\frac { 12 } { 13 } \\tan \\theta = \\frac { B C } { A B } = \\frac { 5 k } { 12 k } = \\frac { 5 } { 12 }{/tex}{tex}\\cot \\theta = \\frac { A B } { B C } = \\frac { 12 k } { 5 k } = \\frac { 12 } { 5 } \\cos e c \\theta = \\frac { A C } { B C } = \\frac { 13 k } { 5 k } = \\frac { 13 } { 5 }{/tex} Sec theata = hypotenuse/ baseSo, base =12 Hypotenuse=13Prependicular =√( hypotenuse)2 - ( base)2 √(13)2-(12)2=√169-144=√25 =5 Sin theata =P/H =5/13Cos theata=B/H =12/13Tan theata=P/B =5/12Cot theata=B/P =12/5Cosec theata= H/P = 13 /5 SecA=H/B,Here secA=13/12So, H=13 and B=12 We know that, P2=H2-B2P2=164-144P2=25P=5Now, All trignometric ratiosSinA=P/H=5/13CosA=B/H=12/13TanA=P/B=5/12CosecA=H/P=13/5CotA=B/P=12/5 |
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| 16368. |
Why sin value of p/h. perpendicular /hypotnese |
| Answer» If you go on Quora digest you got it | |
| 16369. |
-1.2,-3.2,-5.2,-7.2 |
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Answer» It is an AP because the common difference are -2 This is in AP having the common difference of -2 what want to find bo in this question The series decreased by 2 rapidly What it mean bro |
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| 16370. |
Costheta + sintheta = √2costheta prove that costheta - sintheta = √2sintheta |
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Answer» Squaring** Lhs** You can prove it by a square ring on both side and then arrange the lhf in the form of cos theta minus sin theta ka whole square |
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| 16371. |
If TanA+SinA=m and, TanA-SinA=n Then, prove that:- m2-n2=4 under root mn |
| Answer» Squaring both the equation and then subtract each other with as LHS with LHS and RHS with RHS | |
| 16372. |
How i can improve maths |
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Answer» Thanks Perfect practice makes man perfect so by regular practice you can do it By spending more time ? in maths Don\'t solve it simply enjoy it and solve you will improve in maths And regular revision of all theorems and its application By improving your calculation |
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| 16373. |
Let me know to how can i get standard and badic maths questions paper model plz |
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| 16374. |
√2x+√3y=0 |
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| 16375. |
Standard paper model |
| Answer» Khud dhundle mujhe pareshan matkar | |
| 16376. |
Tan70=ta20+2tan50 prove that |
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| 16377. |
Find the roots of the equation x^2 + x -p(p+1) = 0 |
| Answer» If p=1,x= -2 or1 | |
| 16378. |
√5×78 it is rational |
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Answer» No No , it is irrational |
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| 16379. |
Prove that √5 is the irrational |
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Answer» No Lol Raghu Not bad |
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| 16380. |
In maths chapter 4 having completing the square method or nott in board exams |
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Answer» in CBSE board exam 2019 may be completing the square method will not come? What r u asking , your query is not clear I am not understanding you plz say clearly |
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| 16381. |
Tan² theta minus one by cos square theta = -1 |
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Answer» Anjali\'s ans is correct. Consider tan²theta - 1/cos²theta Tan²theta -sec²theta ....(1/cos²theta =sec²theta) identity Therefore tan²theta -sec²theta =-1...(1+tan²theta=sec²theta therefore tan²theta -sec²theta =-1)Hence proved. |
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| 16382. |
How to find the zeros |
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Answer» First factorise the one zero and then the answer is divided by the equation and divided anwer is factorise thrn you have the zerous???? I think you understand By using splitting middle term method and D=b²-4ab formula. Using splitting method and descrimnen(d)=b2-4ac By using middle term splitting and \'D\' method |
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| 16383. |
Tan45.sin60.sec40.cos90 |
| Answer» My Answer is. 0 | |
| 16384. |
Tan30°•sec45°+tan60°•sec30° |
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Answer» How? Root6+6 whole divided by 3 |
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| 16385. |
Explain why 7×11×13+13 and7×6×5×4×3×2×1+5 |
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Answer» a))Take 13 as common in 7×11×13+13=13(7×11+1)=13×78=13×13×6So it has more than 2 factor so it is a composite number.b))Take 5 as common in 7×6×5×4×3×2×1+5=5(7×6×4×3×2+1+1)=5(1008+1)=5×1009Therefor it has more than 2 factor . so it is a composite number. Firstly take 13 common and second 5 solve their equation |
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| 16386. |
441/140 is a rational number with terminating expansion or not |
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Answer» It is a terminating expansion with ans 3.15 Terminating |
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| 16387. |
Prove that 2 is not equal to 0 |
| Answer» 1+1=2where 1 is greater than 0then 2 is more greater than 0Hence,2 is not equal to 0 | |
| 16388. |
value of tan A sq -1 |
| Answer» -sec²A | |
| 16389. |
Sin tan cos ka vastavik |
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| 16390. |
3+2 y = k |
| Answer» Put y= 1 then k=5 and if we put y=2 yhen k=7 | |
| 16391. |
भाज्य bhajya sankhya Kise kehte hain |
| Answer» You ask me | |
| 16392. |
what is quadratic formula ? |
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Answer» -b+_|b^2-4×a×b÷2a _b+√d/2a_b-√d/2a -b±|b^2-4ac/2a |
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| 16393. |
1/cosecA -cot A -1/sin A =1/sin A -1/cosec A +cot A |
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| 16394. |
Crogruence of triangle |
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Answer» Angle sides are equal Angles ,sides is equal |
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| 16395. |
Factorise 12( 4x - 1/4x)^2 + 5(4x - 1/4x) -2 |
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| 16396. |
Ncrt 12.3 exercise |
| Answer» Check\xa0NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 16397. |
Koi ladki reply krna plz |
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Answer» App kon ho Oo snehil ..kitni der baad aaye Hi kyu |
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| 16398. |
Find the sum of odd number between 50 and 100 |
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Answer» 1875 50 Ok Please help! |
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| 16399. |
Degree of p(x) = x + root of(x^2 +1) |
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Answer» 1 1 -1/2 |
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| 16400. |
If angle of elevation of 50 metre high building from a point on the groung is 30 |
| Answer» Please complete the question. | |