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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 7201. |
Prove that 2x^2+4x+3 is a quadratic question.. |
| Answer» The power is 2 | |
| 7202. |
root of 987 |
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Answer» 31.416556145 √987=31.41... |
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| 7203. |
If sec theta + tan theta = p , prove that sin theta = p^2-1/p^2+1 |
| Answer» | |
| 7204. |
1/(2x-3)+1/x-5=1 verify that this is a real root |
| Answer» 1/(2x-3)+1/(x-5)=1(x-5+2x-3)/(2x-3)(x-5)=1(3x-8)/2x2-10x-3x+5=1(3x-8)=2x2-13x+152x2-10x+23=0Here a=2 . b=-10,c=23D=b२-4ac =(-10)२-4(2)(23) =100-148 =-84 D<0Hence,in the given equation has no real root | |
| 7205. |
Whats theorem and how congruency and similarity of two triangles are different ? |
| Answer» | |
| 7206. |
Formula oo trignometry |
| Answer» | |
| 7207. |
How to solve boat and stream question in class 10 maths |
| Answer» By making equation | |
| 7208. |
Djsjaj |
| Answer» Jajsjd | |
| 7209. |
The value of x which satisfies the equation √x^2-4x+3 + √x^2-9 = √4x^2-14x+6 |
| Answer» 7 | |
| 7210. |
What is centeroid |
| Answer» The point where the medians of triangle interest | |
| 7211. |
Ma |
| Answer» | |
| 7212. |
Cot15cot16cot17...cot74cot75 Evaluate |
| Answer» Cot(90-75)cot(90-74).......cot74cot75tan75tan74...cot74cot751 | |
| 7213. |
Write the degree of the polynomial y³ - 2 y² + 3½ y + 7 ????? |
| Answer» 3 | |
| 7214. |
What is quadratic equation |
| Answer» "ax2 + bx + c = 0" is quadratic equation for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor.The Quadratic Formula ax2 + bx + c", where a, b, and c are just numbers; they are the "numerical coefficients" of the quadratic equation they\'ve given you to solve. | |
| 7215. |
Which term of the ap:21,18,15 ........is _81 ? Also is any term o give reason |
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Answer» continue from above check ur answer a35=21+34x-3 = 21-102= - 81 , also a8=21-7x3=21-21=0 ok verified let nth term = - 81a=21d= - 3an=a+(n-1)d- 81=21+3-3nn=105/3=3535th term = - 81also 0=21+3-3nn=24/3=88th term = 0 |
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| 7216. |
if tan9θ=cotθ and 9θ |
| Answer» | |
| 7217. |
If √3tan teetha=3sin teetha,find value of sin square teetha -cos square teetha |
| Answer» | |
| 7218. |
How to solve polynomial |
| Answer» | |
| 7219. |
Plz tell if cbse result for class 10th will include 10 % of preboard 1 and 2 |
| Answer» No I hope | |
| 7220. |
Find the distance between (-5,7), (-1,3) |
| Answer» root of (- 5 +1)2 + (7-3)2= root 16+16= root 32= 4{tex} \\sqrt2{/tex} | |
| 7221. |
1/sec a-1 + 1/sec a+1 = 2 cosec a.cot a |
| Answer» Photo bhejo 9229510263 pe wha5sap | |
| 7222. |
If angle B And angleQ are acute angles such that sinB=sing then prove that angle B =angleQ |
| Answer» Consider two right triangles ABC and PQR in which\xa0{tex} \\angle B{/tex} \xa0and\xa0{tex}\\angle Q{/tex} are the right angles.We have,In\xa0{tex}\\triangle ABC{/tex}{tex}\\sin B=\\frac{AC}{AB}{/tex}\xa0and, In\xa0{tex}\\triangle PQR{/tex}\xa0{tex}\\sin Q=\\frac{PR}{PQ}{/tex}{tex} \\because \\quad \\sin B = \\sin Q{/tex}{tex} \\Rightarrow \\quad \\frac { A C } { A B } = \\frac { P R } { P Q }{/tex}{tex} \\Rightarrow \\quad \\frac { A C } { P R } = \\frac { A B } { P Q } = k{/tex}(say) ...... (i){tex} \\Rightarrow {/tex} AC = kPR and AB = kPQ .....(ii)Using Pythagoras theorem in triangles ABC and PQR, we obtain\xa0AB2 = AC2 + BC2 and PQ2 = PR2 + QR2{tex} \\Rightarrow \\quad B C = \\sqrt { A B ^ { 2 } - A C ^ { 2 } } \\text { and } Q R = \\sqrt { P Q ^ { 2 } - P R ^ { 2 } }{/tex}{tex} \\Rightarrow \\quad \\frac { B C } { Q R } = \\frac { \\sqrt { A B ^ { 2 } - A C ^ { 2 } } } { \\sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = \\frac { \\sqrt { k ^ { 2 } P Q ^ { 2 } - k ^ { 2 } P R ^ { 2 } } } { \\sqrt { P Q ^ { 2 } - P R ^ { 2 } } }{/tex}\xa0[ using (ii) ]{tex} \\Rightarrow \\quad \\frac { B C } { Q R } = \\frac { k \\sqrt { P Q ^ { 2 } - P R ^ { 2 } } } { \\sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = k{/tex}...(iii)From (i) and (iii), we get{tex} \\frac { A C } { P R } = \\frac { A B } { P Q } = \\frac { B C } { Q R }{/tex}{tex} \\Rightarrow \\quad \\Delta A C B - \\Delta P R Q{/tex}\xa0[By S.A.S similarity]{tex} \\therefore \\quad \\angle B = \\angle Q{/tex}\xa0Hence proved. | |
| 7223. |
Trigonometic |
| Answer» | |
| 7224. |
3x-x-4 |
| Answer» 3x-x-4=03x-x=42x=4x=2 | |
| 7225. |
Qartatic equation |
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Answer» Equation in with highest degree is t w o is known as quadratic equation Tuze nahi ata |
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| 7226. |
Find the sum of 51 terms of the A. P. whose second term is 2 and the fourth term is 8. |
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Answer» 1,2,3,4,5,6......a+d=2 . a=2-d ................. ia+3d=8 .........iifrom i and ii2 - d +3d=82d=6d=3a= - 1S51 = 51/2(2a+ (51-1)d) = 51a+25x51x3 = 51x-1 + 3825 = 3774\xa0 a+d=2a+3d=8-d=-6d=6a+6=2a=-4 |
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| 7227. |
Find the sum of first 100 even natural numbers which are divisible by 5.This is a question of A. P. |
| Answer» numbers are 10,20,30,40,........................they form an APa=10d=10s100 = 50(20 + 990) = 50 x 1010 = 50500 | |
| 7228. |
Sa 1 questions paper please |
| Answer» | |
| 7229. |
2x+4x+3y=0 |
| Answer» =6x+3y=0 2x+y=0When x=1 then Y=-2When x=-3. Then y=-6 | |
| 7230. |
Find the 6 term from the end of the AP 17,14,11---- -40 |
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Answer» Here a=17 d=14-17=-3Tn=-40a+(n-1)d=-4017+(n-1)-3=-4017-3n+3=-4020-3n=-40-3n=-40-20-3n=-60n=60/3n=20Total number of term in this AP=206th term from the end =(20-6)+1=15th terms T15=a+14d=17+14×(-3)=17-42=-256th term is - 25 zpwdjjqaxjssiw23 |
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| 7231. |
How to do *** |
| Answer» | |
| 7232. |
X3+3x2+3x+1÷2x+1 plz answer only |
| Answer» | |
| 7233. |
(a-b) cube |
| Answer» a3-b3-3ab(a-b) | |
| 7234. |
I want the solution of sample paper 2008-09 half yearly...Plz someone help me to get the solution... |
| Answer» | |
| 7235. |
TheoremArea base |
| Answer» | |
| 7236. |
Write down the prime factrosation of 720 |
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Answer» 2*2*2*2*5*3*3 2*2*2*2*5*3*3 |
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| 7237. |
Tan square theata + cot square theta +2 = SEC square theata * cosec squre theata |
| Answer» We have,{tex} \\mathrm { LHS } = \\frac { \\tan \\theta - \\cot \\theta } { \\sin \\theta \\cos \\theta } = \\frac { \\frac { \\sin \\theta } { \\cos \\theta } - \\frac { \\cos \\theta } { \\sin \\theta } } { \\sin \\theta \\cos \\theta } = \\frac { \\frac { \\sin ^ { 2 } \\theta - \\cos ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta } } { \\cos \\theta \\sin \\theta }{/tex}{tex} \\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\sin ^ { 2 } \\theta - \\cos ^ { 2 } \\theta } { \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta }{/tex}{tex} \\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\sin ^ { 2 } \\theta } { \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta } - \\frac { \\cos ^ { 2 } \\theta } { \\sin ^ { 2 } \\theta \\cos ^ { 2 } \\theta }{/tex}{tex} \\Rightarrow \\quad \\mathrm { LHS } = \\frac { 1 } { \\cos ^ { 2 } \\theta } - \\frac { 1 } { \\sin ^ { 2 } \\theta }{/tex}{tex} \\Rightarrow \\quad L H S = \\sec ^ { 2 } \\theta - \\ cosec ^ { 2 } \\theta = \\left( 1 + \\tan ^ { 2 } \\theta \\right) - \\left( 1 + \\cot ^ { 2 } \\theta \\right) = \\tan ^ { 2 } \\theta - \\cot ^ { 2 } \\theta = \\mathrm { RHS }{/tex} | |
| 7238. |
Formula for the midpoint of class interval |
| Answer» | |
| 7239. |
Ncert pg-76 Qu no-2 |
| Answer» By spilleting method for ex. The midle term of qus no.1when we add and subtract the given no. Is 3 and we multiply the no.is.10 | |
| 7240. |
If sinx+sin^2x+sin^3x=1Then 4cosx=? |
| Answer» | |
| 7241. |
Radius =150 , circumference? |
| Answer» Given,Radius =150We know that,Formula of circumference if 2πrNow,=2×3.14×150=6.28×150=942Hope this ans will be helpful to you.. | |
| 7242. |
formulae of median |
| Answer» Median =3mean-2mode maybe | |
| 7243. |
Is optional question are coming in examination of subject math or other side books |
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Answer» Refer ncert maths text book pg 171 last line. No.. |
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| 7244. |
Which type of questions bank is important for board exam |
| Answer» | |
| 7245. |
In class 10 mainly exam comes from ncert or extra¤¤¤¤ |
| Answer» | |
| 7246. |
Is there will be internal choice will be given in class 10 board paper of Mathematics |
| Answer» No.......there no choices in CBSE maths | |
| 7247. |
Find the value of cos theta and sin theta |
| Answer» 1 | |
| 7248. |
44+55=33 kaise hoga |
| Answer» Not possible | |
| 7249. |
Jwyjxjhd |
| Answer» | |
| 7250. |
If sin Q and cos Q are roots of equation (2p+1)x^-7p-3=0 has equal roots |
| Answer» The given quadratic equation is:\xa0(2p + 1)x2 - (7p + 2)x + 7p - 3 = 0Here, a = (2p + 1), b = -(7p + 2) and c = 7p - 3We know that, D = b2 - 4ac= [-(7p + 2)]2 - 4 {tex}\\times{/tex} (2p + 1) {tex}\\times{/tex} (7p - 3)= 49p2 + 4 + 28p - 4(14p2 - 6p + 7p - 3)= 49p2 + 4 + 28p - 4(14p2 + p - 3)= 49p2 + 4 + 28p - 56p2 - 4p + 12= -7p2 + 24p + 16Since it is given that the given equation has real and equal roots, so D = 0 i.e.,{tex}\\Rightarrow{/tex}\xa0-7p2 + 24p + 16 = 0{tex}\\Rightarrow{/tex}\xa07p2 - 24p - 16 = 0{tex}\\Rightarrow{/tex}\xa07p2 + 4p - 28p - 16 = 0{tex}\\Rightarrow{/tex}\xa0p(7p + 4) - 4(7p + 4) = 0{tex}\\Rightarrow{/tex}\xa0(p - 4)(7p + 4) = 0{tex}\\Rightarrow{/tex}\xa0p - 4 = 0 or 7p + 4 = 0{tex}\\Rightarrow{/tex}\xa0p = 4 or {tex}p = - \\frac{4}{7}{/tex}Therefore, after substituting the value of p in the given equation, the two equations will be:9x2 - 30x + 25 = 0 [For p = 4]{tex}\\Rightarrow{/tex}\xa09x2 - 15x - 15x + 25 = 0{tex}\\Rightarrow{/tex}\xa03x (3x - 5) - 5x (3x - 5) = 0{tex}\\Rightarrow{/tex}\xa0(3x - 5)2 = 0{tex}x = \\frac{5}{3}{/tex}or\xa0{tex}\\left( {2 \\times \\left( { - \\frac{4}{7}} \\right) + 1} \\right){x^2} - \\left( {7\\left( { - \\frac{4}{7}} \\right) + 2} \\right)x{/tex}{tex} + 7\\left( { - \\frac{4}{7}} \\right) - 3 = 0{/tex}\xa0[For {tex}p = - \\frac{4}{7}{/tex}]On solving we get,x2 - 14x + 49 = 0{tex}\\Rightarrow{/tex}\xa0x2 - 7x - 7x + 49 = 0{tex}\\Rightarrow{/tex}\xa0x (x - 7) - 7 (x - 7)= 0{tex}\\Rightarrow{/tex} (x - 7)2\xa0= 0{tex}\\Rightarrow{/tex} x = 7Thus,\xa0{tex}x = \\frac{5}{3}{/tex} or\xa0x = 7 | |