InterviewSolution
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InterviewSolution
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.
| 1151. |
What do you mean by quadratic formula?How can you explain the nature of roots ????? |
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Answer» Hii avi Hi jagrati ???? Which school do you read??Drift Bøî Thank you Use the quadratic formula to determine the roots of the quadratic equations given below and take special note of:the expression under the square root sign andthe type of number for the final answer (rational/irrational/real/imaginary) and we find that those ans are 1.equal to 0 ( 2.less then 0 3. And more then 0 |
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| 1152. |
Find the value of \'a\' so that the point (3,9) lies on the represented by 2x-3y=5 |
| Answer» It is given that ,( 3 , a ) lies on the line 2x - 3y - 5 = 0Substitute x = 3 , y = a in the equationWe get ,2 × 3 - 3a - 5 = 06 - 3a - 5 = 01 - 3a = 01 = 3a1/3 = aa = 1/3 | |
| 1153. |
(Sin3A+cos3A)/sinA+cosA + sin3A - cos3A/sinA+cosA=2 |
| Answer» ??? | |
| 1154. |
2 x square - 3 x 5 is equal to zero |
| Answer» 2x² -3*5 =02x² = 15x² = 7.5x= √7.5 =2.74 or -2.74 (approx.) | |
| 1155. |
10+10+30+10+90+11+23+35+378+679×6893+3784+63+÷63+579×4578÷245+×3476×4 |
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Answer» 8823976.029 8823976.029 |
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| 1156. |
Holiday home work of linear equations in one variable |
| Answer» | |
| 1157. |
G(x) = x ki power 3 +3x and p(x) x ki power 2 + 3xki power 4+ 2x |
| Answer» | |
| 1158. |
What is sinA.sinB |
| Answer» Dear that\'s not in ur syllabus | |
| 1159. |
Tell me thearom 6.3 |
| Answer» the congruency of a triangle, if two sides of a triangle and angle between them are equal, both the triangles are congruent to each other. | |
| 1160. |
If sin A=3/4 caluculate cos A and tan A |
| Answer» CosA=5/4,tanA=3/5 | |
| 1161. |
1-60×1/80/15 |
| Answer» 1-60x1/80x15=1-1/20=20-1/20=19/20 ans__ | |
| 1162. |
If tan theta + cos theta equal 2 find sec square theta into cosec square theta |
| Answer» | |
| 1163. |
If angle A and Angle B are acute angle, where CosA=CosB. Show that Angle A= Angle B |
| Answer» A/q,CosA=CosB => AC/AB=BC/AB => AC=BC . Here,we know that any triangle whose two adjacent side are equal then it\'s corresponding angle are also equal. So, angleA=angle B. | |
| 1164. |
In which topic should I make Blogs? |
| Answer» Blog is a type of social adventure in this u should blog on any yopic like poorness , animals ,gagets ,places and many things. | |
| 1165. |
I am good cricketer can i continue study after getting chance in ipl |
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Answer» Yes u can......... ☺ If u wish u can ..study never ends If u think that u r having the capability to play ipl nd if your parents r not having any prblm with ur decision.......so why not u can continue your study after getting chance in ipl..... |
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| 1166. |
I am a cricketer i play cricket so how i get better result in board exam |
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Answer» Leave about cricket for this year During study time don\'t think about ? cricket and try to focus on your study?????? Make fix time of studying and playing |
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| 1167. |
How many solutions does the pair of equation y=0 and y=-5 have ? |
| Answer» 1 solution | |
| 1168. |
If 3sin theta equal 4cos theta.find the value of 5cos theta+7 upon 5 cos theta - 7 |
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| 1169. |
Sin 60 cos 30+sin 30 cos 60 |
| Answer» √3/2×√3/2+1/2×1/2 => 3/4+1/4=> 1 | |
| 1170. |
If cos 9a = sin a, then the value of a is |
| Answer» Sin(90°-9a)=sin a => 10a=90° => a=9° | |
| 1171. |
a2x2+3abx+2b2Find the roots |
| Answer» a²x² - 3abx + 2b² = 0a²x² - 3abx = - 2b²divide each term with a² both sides of theequation ,x² - 3bx/a = - 2b²/a²x² - 2 × x × (3b/2a)+(3b/2a)² = ( 3b/2a )² - 2b²/a²( x - 3b/2a )² = 9b²/4a² - 2b² /a²= ( 9b² - 4b² ) /a²= 5b²/a²x - 3b/2a = ± √ ( 5b²/a² )x = 3b/2a ± √5 ( b/a )= [ 3b ± 2√5 b ] / 2a= ( 3 ± 2√5 )b/ 2a(I hope this helps you.)??? | |
| 1172. |
Who made this maths sub ? |
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Answer» Those who r tolerating it ( humans).. Maths is the invention of humans Beginning in the 6th century BC with the Pythagoreans, the Ancient Greeks began a systematic study of mathematics as a subject in its own right with Greek mathematics. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof. |
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| 1173. |
3X2-7X+3=0 |
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Answer» Thanks X=-(-7)±√13/2×3 => 7±√13/6 Thank u D=(-7)²-4×3×3=> 49-36=> D=13 |
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| 1174. |
Exercise 8.2 |
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Answer» I know it kasht karlenge hum Very nice reply pranali nd yss its available on this app....? Ya you are right pranali the exercise 8.2is available in the app? Available in this app ?plz?thoda kasht kre ? |
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| 1175. |
1000000000000-10000000000 |
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Answer» 9.9E11 ? |
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| 1176. |
2^36 – 1 = 68a19476735 then the value of ‘a’ is ----------------------------- . |
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| 1177. |
If f(x)=ax4+bx3+cx2+dx+ e is apolynomial of degree 4having a,B¥and 8as it s zeros ? |
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| 1178. |
Show that there is no positiv integer n for which √ n-1+√n+1is ratinol |
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| 1179. |
If a,b,c are in AP then prove that, (bc-a²),(ca-b²),(ab-c²) are in AP. |
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Answer» Koi na Given: a, b and c ar in APAs we know, if a, b and c are in AP then 2b = a + c------ (1)We have to prove that ( bc - a2\xa0), ( ca - b2\xa0), ( ab - c2\xa0) are in AP.For this we should prove : 2( ca - b2\xa0) = ( bc - a2\xa0)+ ( ab - c2\xa0)Let us consider\xa0( bc - a2\xa0)+ ( ab - c2\xa0)= ab + bc - (a2+c2)= b( a + c ) - [ (a + c)2\xa0- 2ac] [\xa0∵\xa0a2+b2\xa0=\xa0(a + b)2\xa0- 2ab\xa0]= b(2b) - [ ( 2b )2\xa0-2ac] [\xa0∵ From (1) ]= 2b2\xa0- 4b2\xa0+ 2ac= 2ac -\xa02b2= 2(ca - b2) Don\'t know sorry? |
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| 1180. |
58+5588668 |
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Answer» 5,588,726 5588726 is answer of this que. Seriously #daytillworldend |
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| 1181. |
Find the sum of all integers from 1 to 2001 |
| Answer» Sn=2000/2(1+2001)=1000×2002=2002000 | |
| 1182. |
Show that any number of the form 6n,where n€N can never end with the digit 0. |
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Answer» Thanx ?????❤? Taking some examples of the numbers ending with digit 0 like 10=2×5,20=2×2×5,30=2×5Here we note that numbers ending with digit 0 has both 2 and 5 as their prime factorsWhereas prime factors of 6n is (2×3)n and fire is not in the prime factorization of 6n so it cannot be end with digit 0I hope you will understand |
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| 1183. |
But R. D. is very long and exemplar is good |
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Answer» Solving NCERT maths and exemplar with honesty is enough.Other reference books are-1.RS Agarwal2.All in one3.S chand mathematics You should try RS Aggarwal or All in one...☺️☺️ |
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| 1184. |
If the point (p+1, p), (p, p+1), (p-1, p-2)are collimeart then find the value of p |
| Answer» Slope between (p+1,p), (p, p+1) = slope between (p, p+1), (p-1,p-2) (p+1-p)/[p-(p+1)] = [(p-2)-(p+1)]/(p-1-p)1/-1 = -3/-1-1 = 3 | |
| 1185. |
Find the circumcenter of a circle passing through the points 6, -6&3,-7&3,3 |
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| 1186. |
Find the centre of a circle passing through the points 6, - 6 and 3, - 7 and 3, 3 |
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| 1187. |
Ex:7.4 3rd question |
| Answer» Find the centre of circle passing through the points 6, - 63, - 7 and 3, 3 | |
| 1188. |
√(x/2) = 3 |
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Answer» In simple way, x/2=3² => x=9×2 => x=18 Start with:√(x/2) = 3Square both sides:x/2\xa0= 32Calculate 32\xa0= 9:x/2 = 9Multiply both sides\xa0by 2:x = 18 |
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| 1189. |
What is the Euclid division algorithms |
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Answer» a=bq +r, where r=,or less than b Euclids division algorithm states that for every two positive integers a and b there unique exist q and r such that a=bq+r,where 0 |
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| 1190. |
Tan^2 A/tan^2A -1 + cosec^2A / sec^2A - cosec^2 A = 1 / 1-2cos^2A |
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| 1191. |
Practice paper solutions chapters 4 quadratic equations |
| Answer» Available in this app?plz?thoda ksht kre ? | |
| 1192. |
Chapter 4 quadratic equations practice paper solutions |
| Answer» Available in this app ?plz ? thoda ksht kre? | |
| 1193. |
Find the nature of the roots of the eqn 4x2-12x-9=0Any help ,then thankyou? |
| Answer» D=(-12)²-4×4(-9) => 144+144= 288. So it\'s nature is real and equal | |
| 1194. |
solve by reducing method 4 by x + 3y = 14 and 3 by x - 4y = 23 |
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Answer» 4/x + 3y=14 and 3/x - 4y=23Let 1/x = m,4m + 3y = 14...(1)and 3m - 4y = 23 ... (2)Multiply (1) by 4 and (2) by 3, we get16m + 12y = 56 ... (3) and 9m - 12y = 69 .... (4)Adding (3) and (4), we get25m = 125→ m = 5From (1),4(5) + 3y = 14→ 3y = 14 - 20→ y = -2\xa0and m = 1/x = 5 → x = 1/5 x=1/5, y=-1 |
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| 1195. |
10x-1/x=3,x=0 |
| Answer» 10×3 (x=3)10×3-1=2910×0(x=0)10×0-1=-1 | |
| 1196. |
Find the smallest number which when divided by 30 and 40 and 60 leaves the remainder 7 in each case |
| Answer» 127 | |
| 1197. |
Find the smallest number which when divided by 30 40 and 60 |
| Answer» 10 | |
| 1198. |
The 8 th term of AP is -23 and its 12 term is -29find the AP |
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Answer» Right answer Let the first term be a and common difference be d. A/q a+7d=-23 ....(१) and a+11d=-२9....(२) equate both equation then, d=-3/2 and first term is -67/2 |
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| 1199. |
How many terms of the AP 26,21,16,11,.... are needed to give the sum 11? |
| Answer» \'a\' or the first term=26and \'d\' or the common difference =21-26=-5n=?Sum of n terms of A.P=11Sum of n terms of A.P=n/2(2a+(n-1)d11=n/2(52-5n+5)22=n(57-5n)22=57n-5n25n2-57n+22=0By splitting the middle term,5n2-55n-2n+22=05n(n-11)-2(n-11)=0(5n-2)(n-11)=0n-11=0n=11 | |
| 1200. |
1hour contai how many seconds |
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Answer» Hi Koi na ? Thanks Seconds 3600 |
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