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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.
| 61201. |
(11)}=(-9)6*263)e following |
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Answer» -6/35-1/4-1/35=-7/35-1/4=-1/5-1/4=-9/20 |
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| 61202. |
10anges of a quadrilateral aree following can bel Two angles |
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| 61203. |
5. Find the value of38o(-1)" + (-1)2n + (-1)positive odd integer.21+1+ (-1)4n+2 ; where n is any |
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| 61204. |
If n is an odd integer, then show that ? - 1 is divisible byProve that |
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Answer» Put the value of n =8n+1, 8n+3, 8n+5 x^2-1=8; x^2=9; x=3 |
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| 61205. |
If n is an odd integer then show that n2-1 is divisible by 8. |
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Answer» Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p.Let n = 4p+ 1,(n^2 – 1) = (4p + 1) ^2– 1 = 16p^2+ 8p + 1 = 16p^2+ 8p = 8p (2p + 1) (n2– 1) is divisible by 8.(n2– 1) = (4p + 3)^2– 1 = 16p^2+ 24p + 9 – 1 = 16p^2+ 24p + 8 = 8(2p^2+ 3p + 1) n^2 – 1 is divisible by 8. Therefore, n^2 – 1 is divisible by 8 if n is an odd positive integer. |
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| 61206. |
if n is an odd positive integer then show that (n^2 - 1) is divisible by 8. |
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Answer» If Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p.Let n = 4p+ 1,(n^2– 1) = (4p + 1)^2– 1 = 16p^2+ 8p + 1 = 16p^2+ 8p = 8p (2p + 1)⇒ (n^2– 1) is divisible by 8.(n^2– 1) = (4p + 3)^2– 1 = 16p^2+ 24p + 9 – 1 = 16p^2+ 24p + 8 = 8(2p^2+ 3p + 1)⇒ n^2– 1 is divisible by 8.Therefore, n^2– 1 is divisible by 8 if n is an odd positive integer. |
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| 61207. |
what is meant by parameters and percentages |
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Answer» Parameter :Aparameteris a quantity that influences the output or behavior of amathematicalobject but is viewed as being held constant.Parametersare closely related to variables, and the difference is sometimes just a matter of perspective. Percentage Apercentis a ratio whose second term is 100.Percentmeans parts per hundred. The word comes from the Latin phrase per centum, which means per hundred. Inmathematics, we use the symbol % forpercent. thanks |
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| 61208. |
के... B सर 309 22: 2 कि (67 छा कक धरके... नितिन1 दर 00 110 3-2. 153 (५5 स्व ९ (1. हद—IR |
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| 61209. |
95 65 24 172Example 3 Find the sum of andSolution:5 \frac{95}{100}, \frac{65}{52}, \frac{24}{33} \text { and } \frac{172}{110} |
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| 61210. |
3+7(10 x14) |
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| 61211. |
log(2*x %2B 3, 10)=log(7, 10) |
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Answer» Kon sa chapter ka he |
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| 61212. |
(a) If α and β are the zeros of the quadratic polynomial p(1)2-5-1, find the value of |
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Answer» it's not a correct answer ok |
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| 61213. |
) If α and β are the zeros of the quadratic polynomial p(i)pr-1, find the value or |
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Answer» first read question properl |
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| 61214. |
12. Using factor theorem, show that z + a is a factor of *'+ a' for any odd integer n. |
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| 61215. |
-5x-1, find the valu9.If α andα and β are the zeros of the quadratic polynomial p(x) = 4x |
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| 61216. |
123. दी गई भुजाओं की मापों में से किसमाप से त्रिभुज सम्भव है ?(&) 2 सेमी, 3 सेमी, 5 सेमी(8) 6 सेमी, 3 सेमी, 2 सेमी(0) 3 सेमी, 6 सेमी, 7 सेमी(09) 4 सेमी, 5 सेमी, 9 सेमी |
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Answer» answer c. yoga. kiyoki. C is the answer because sum of any two sides of triangle is always greater than the third side |
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| 61217. |
(00 छिप जद sqpans g ‘का infegeq i ।e o0 st & vé_fi& <Orne inteoes |
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Answer» Like if you find it useful is it a correct one |
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| 61218. |
-09%-५ड0०- : 5 |
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Answer» x^2-x+4x-4=0x(x-1)+4(x-1)=0(x-1)(x+4)=0x=1,-4 |
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| 61219. |
Write each of the following as decimals:(a)(b) 3710(c) 200 +60 + 5 + 1(d) 70+101012 s(0 41 (g)10 |
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Answer» (a)0.5(b) (30+7)/10 = 3.7c) 265+0.1= 265.1d) 70+0.8= 70.8e)42/10= 4.2f)1.5 |
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| 61220. |
GFind the sopplemet of each of the following angles:00 95() 110n) 115(iv) 1351. |
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Answer» Supplement of angle x is given by 180-x i) 180-95 = 85ii) 180-110 = 70iii) 180-115(1/2) = 64(1/2)iv) 180-135(1/2) = 44.5 |
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| 61221. |
factories x^2 +z^2 - 2xz |
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Answer» x^2 + z^2 - 2xz = x^2 - xz - xz + z^2 = x(x - z) - z(x - z) = (x - z)(x - z) |
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| 61222. |
Q31 Using factor theorem factories: x '+13x +32x+20 |
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Answer» x*x*x + 13x*x + 32x + 20 = x*x*x + x*x + 12x*x + 12x + 20x + 20 = (x+1)(x*x+12x+20) = (x+1)(x+2)(x+10) If you find this answer helpful then like it. |
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| 61223. |
031 Using factor theorem factories: x'+13x +32x+20 |
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| 61224. |
09 5. oJ0X 60 |
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Answer» 8,200. sari naaa reeeeee |
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| 61225. |
"POOOOOOOO900PONTON5. Find a quadratic polynomial, the sum and product of whose zeroes areand - 3 respectively.Annly the division alcorithm to find the quotient and remainder on dividing niya 4 _ 3x2 + 4 |
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| 61226. |
\frac { \operatorname { cot } \theta - \operatorname { cos } \theta } { \operatorname { cot } \theta + \operatorname { cos } \theta } = \frac { \operatorname { cosec } \theta - 1 } { \operatorname { cosec } \theta + 1 } |
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Answer» now sintheta=1/cosecthetahence rhs will.be cosectheta-1/cosectheta+1 |
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| 61227. |
11.If α and β are the zeroes of the quadratic polynomial p(s)-3s^2-6s+ 4, then the value of \frac{\alpha}{\beta}+\frac{\beta}{\alpha}+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3 \alpha \beta(A) 6(B) 8(C) 5(D) 4 |
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Answer» alpha + beeta = -(b/a) alpha + beeta = -(b/a) |
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| 61228. |
If α and β be the roots of the equation 3x^2-6x + 40, find the value of\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+3 \alpha \beta |
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| 61229. |
\frac{\sin \alpha}{\sin \beta}-\frac{\cos \alpha}{\cos \beta}=\frac{\sin (\alpha-\beta)}{\sin \beta \cdot \cos \beta} |
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| 61230. |
a ( \frac { \alpha ^ { 2 } } { \beta } + \frac { \beta ^ { 2 } } { \alpha } ) + b ( \frac { \alpha } { \beta } + \frac { \beta } { \alpha } ) |
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Answer» Question you have submitted is incomplete. Please post a complete question. 8th one |
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| 61231. |
$ \sqrt{\frac{1+\sin \theta}{1-\sin \theta}} $ is equal to(a) $ \sec \theta+\tan \theta $(c) $ \sec ^{2} \theta+\tan ^{2} \theta $ |
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Answer» option (a) is correct |
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| 61232. |
\left| \begin{array}{ccc}{\cos \alpha \cos \beta} & {\cos \alpha \sin \beta} & {-\sin \alpha} \\ {-\sin \beta} & {\cos \beta} & {0} \\ {\sin \alpha \cos \beta} & {\sin \alpha \sin \beta} & {\cos \alpha}\end{array}\right| का मान ज्ञात कीजिए।sina cosp sino sinf cosa |
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| 61233. |
Ifa, β are the zeroes of a quadratic polynomial 3x2-6x + 4, find the value of |
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| 61234. |
2xsquare minus 7x minus 15 |
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Answer» Ans :- 2x^2 - 7x -15 = 2x^2 -10x +3x -15 = 2x(x-5) +3(x-5) = (x-5)(2x +3) PLEASE LIKE THE SOLUTION |
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| 61235. |
9. If y = x + x2 + x3 +09, where | x | < 1, then prove that x = |
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Answer» y = x+x²+x³ ...... is a g.p having first term a = x and r= x , so, sum of this g.p is = y and also sum.of g.p , when common ratio is < 1 is S = a/(1-r) here it is y = x/(1-x)=> y(1-x) = x => y-yx = x => y = x+yx=> y = x(y+1)=> x =y/(y+1) |
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| 61236. |
09. Factories x + xy +8x +8 |
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Answer» x^2 + xy + 8x + 8y = x(x + y) + 8(x + y) = (x + 8)(x + y) |
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| 61237. |
Find matrices X and Y if:5 209 and X - Y-3 610 -1 |
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| 61238. |
If the sum of the zeroes of the quadratic polynomial 3x2 - kx + 6 is 3, then find k |
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| 61239. |
2Q13. If one zero of the polynomial 3x2 - 8x + 2k +1 is seven times the other find the zeros and k. |
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Answer» thanks |
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| 61240. |
Write the IUPAC name of given compound |
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| 61241. |
one upon sec theta minus tan theta equal to sec theta + tan theta prove that |
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| 61242. |
13In how many ways can 7 plus (+) signs and 5minus (-) signs be arranged in a row so that notwo minus signs are together? |
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| 61243. |
\left. \begin{array} { l } { 2 \operatorname { sin } ^ { 2 } \beta + 4 \operatorname { cos } ( \alpha + \beta ) \operatorname { sin } \alpha \operatorname { sin } \beta + \operatorname { cos } 2 ( \alpha + \beta ) = } \\ { ( a ) \operatorname { sin } 2 \alpha } \\ { ( c ) \operatorname { cos } 2 \alpha } & { ( d ) \operatorname { sin } 2 \beta } \end{array} \right. |
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| 61244. |
\frac { \operatorname { cosec } \theta + \operatorname { cot } \theta - 1 } { \operatorname { cot } \theta - \operatorname { cosec } \theta + 1 } = \frac { 1 + \operatorname { cos } \theta } { \operatorname { sin } \theta } |
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| 61245. |
\frac { \operatorname { cot } \theta + \operatorname { cosec } \theta - 1 } { \operatorname { cot } \theta - \operatorname { cosec } \theta + 1 } = \frac { 1 + \operatorname { cos } \theta } { \operatorname { sin } \theta } |
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Answer» Put theta = aCot a-1+cosec a/cot a +1-cosec a= (cot a+cosec a)-1/cot a-cosec a+1=cot a +cosec a- (cosec^2a-cot^2a)/cot a - cosec a+1=(cot a+cosec a)-(cot a+cosec a)(cosec a-cot a)/cot a-cosec a+1=(cot a+cosec a)[1-(cosec a-cot a)]/cot a-cosec a+1=(cot a+cosec a)(1+cot a-cosec a)/cot a- cosec a+1=cot a +cosec a=(cos a/sin a)+1/sin a=1+cos a/sin a |
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| 61246. |
solve thefollowingequations byusing eliminatiormethod㥠minus6 y I equal to2ero and x minusequal to zero |
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Answer» 5x-6y-1=0x-12(y+1)/5=0 5x-6y=1------(1)5x-12y=12------(2)eqⁿ1-26y=-11y=-11/6 put in eq15x-6(-11/6)=15x+11=15x=-10x=-2 |
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| 61247. |
ed the compound interest and the amount on 72500! Find the coat 8 %for 2 years. |
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| 61248. |
3x upon 5 minus 8 is equal to minus 9 |
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| 61249. |
3. Find the value of k, when (x- 5) is a factor of thepolynomial 3x2+kx-10. |
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| 61250. |
simplify X by 3 + 1 by 5 - X by 3 minus 1 by 5 |
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Answer» x/3+1/5-x/3-1/5All terms are cancelled=0 |
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