Saved Bookmarks
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.
| 11151. |
3x-8y=9 |
| Answer» | |
| 11152. |
5x+4+=7 |
| Answer» 5x=7-45x=3x=3/5 | |
| 11153. |
3/2 |
| Answer» | |
| 11154. |
Prove BPT THEOREM |
| Answer» In ncert book triangles 1 st theorem | |
| 11155. |
Kyaa isme thurems nahi hai |
| Answer» | |
| 11156. |
What is Basic Proportionality Theorem |
| Answer» If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio. | |
| 11157. |
6.2 |
| Answer» | |
| 11158. |
If sin^25=sinθ, then find the possible values of θ |
| Answer» It\'s 25 | |
| 11159. |
A rectangle having sides x+3y,7,13,3x+y.Find the value of x and y. |
|
Answer» We know Opposite sides of a rectangle are equal So, x+3y=13.......(1)3x+y=7.............(2)By elimination method Equation (1) multiply by 3 So , 3x+9y=39 3x+ y=7 (on subtraction)We give, y=4 Now ,put the value of y in equation (1) X+3y =13 X+3(4)= 13 X=13-12 X=1 Y=1 and x=4 |
|
| 11160. |
Sin 10°= |
| Answer» | |
| 11161. |
Find the roots of the equation 5xsquare minus 6x minus 2=0 |
| Answer» Use completing square method | |
| 11162. |
what is a minus b whole square |
|
Answer» b square Plus b square A^2+b^2-2ab +bsquare +b Plus b |
|
| 11163. |
mth =n nth =m prove pth =m+n-p |
| Answer» m+n-p | |
| 11164. |
If a,b,c, are in A.P. then (a-c)^2 ÷ (b^2-ac) |
| Answer» | |
| 11165. |
Cosec^2 theta + sec^2 theta = cosec^2 theta×sec^2 theta |
| Answer» 1/sin^2 theta+1/cos^2thetaSin^2theta+cos^2theta/sin^theta×cos^theta1/sin^2theta×cos^2thetaCosec^2theta×sec^2theta | |
| 11166. |
Trigometry |
| Answer» TRIGONOMETRY MEANS...TRI WHICH MEAN THREE , GONO MEAN SIDES AND METRY MEAN MEASUREMENT...SO, THREE SIDES MEASUREMENT IS CALLED TRIGONOMETRY | |
| 11167. |
Formulae of CH 7 |
| Answer» Check formulae here :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 11168. |
If x+1÷x =5 ten find the value of x^51+1÷x51 |
| Answer» | |
| 11169. |
HloWhat will be the scheme and division of mark Chapter ise in sa-1 |
| Answer» | |
| 11170. |
Sin^2A +3cos^2A=4 show that tan A=√3÷1 |
| Answer» | |
| 11171. |
Cos90 |
|
Answer» 0 cos 900 = 0 |
|
| 11172. |
Prove that cos^2 theta+cos^2 theta×cot^2 theta=cot^2 theta |
|
Answer» cos2Ɵ + cos2Ɵ x cot2Ɵ = cot2ƟLHS=taking\xa0cos2Ɵ common we get cos2Ɵ (1+\xa0cot2Ɵ) = cos2Ɵ x cosec2Ɵ =\xa0cos2Ɵ x 1/ sin2Ɵ =\xa0cot2Ɵ = R.H.S COS^2 THETA + COS^2 THETA X COT^2 THETA = COS^2 THETA + COS^2 THETA X COS^2THETA /SIN^2THETA (COT^2 THETA =COS^2 THETA/SIN^2 THETA)THEN, (COS^2 THETA X SIN^2 THETA + COS^4 THETA )/SIN^2THETA =COS^2 THETA (SIN^2 THETA + COS^2 THETA)/SIN^2 THETA (TAKING COS^2 THETA Common) =COS^2 THETA (1)/SIN^2 THETA (BECAUSE COS^2 THETA + SIN^2 THETA = 1 ) = COT^2 THETA. Hence Proved |
|
| 11173. |
2ax+3by=a+2b3ax+2by=2a+bSolVe this by croSs mulTipLicaTion..metHoDd.. |
| Answer» 2ax + 3by = a + 2b⇒\xa02ax = a+2b – 3by⇒\xa0x = (a+2b-3by) / 2aPut x in the second equation3ax + 2by = 2a+b⇒ 3a((a+2b-3by) / 2a) + 2by = 2a+bBy solving it,\xa0we get\xa0y = (4b-a)/5bx = (a+2b-3by) / 2a\xa0x = (4a-b)/5a | |
| 11174. |
Prove that √3 is irrational. |
|
Answer» I have alredy given answer for this question Let us take root 3 as rational number The rational number is of the form a/b where a, b are integers and b not equal to zero, such that a is equal to root 3 bA square is equal to 3 b square Therefore a square is divisible by 3That implies A is also divisible by 3Let us take b is equal to 3kA square is equal to 3 into 3 k square3a squareis equal to 3 into 3 k squareTherefore b square is divisible by 3That implies B is also divisible by 3A and b has a common factor 3Bat this contradicts to the fact that a/b is irrational That implies root 3 is irrational |
|
| 11175. |
99x+101y=499 and 101x+99y=501Solve it?Plz suggest a trick to solve this quickly |
|
Answer» Let us use a trick to solve it quickly99x + 101y = 499 ............ i101x + 99y = 501............ iisubtract i from ii2x - 2y = 2 or x - y =1 or x = 1 + y ....... .............................................. iiiadd i and ii200x + 200y = 1000 or x + y = 5 ...................................................................... ivNow from iii and iv , we get 1 + y + y = 5 or 2y = 4 or y = 2 ...... vagain from iii and v , we get x = 1 + 2 = 3Hence x = 3 , y = 2 99x+101y=499 .....(1)101x+99y=501 ......(2)Add (1) and (2), we get200x + 200y = 1000x+y = 5 ...... (3)Subtract (1) from (2), we get2x -2y = 2x - y = 1........ (4)Add (3) and (4), we get2x = 6=> x = 3Put value of x in (3), we get3+y = 5=> y = 2\xa0\xa0\xa0 |
|
| 11176. |
For 5th standard mathematics |
| Answer» | |
| 11177. |
How we proof √2+5 is an irrational number. |
|
Answer» Let √2+5 is a rational number.we take 5 another rational number.We know that difference of two rational number is rational number.Here difference of both number is √2.But it is an irrational number.Hence our supposition was wrong.So √2+5 is an irrational number.Proved. Let √2+5 be a rational numberThen,. √2+5=a/b. Where a and b are integers and b is not equal to zero √2=5+a/bAs we know that sum of two rational number is always a rational |
|
| 11178. |
What does CPST mean? |
|
Answer» Corresponding Part of Similar Triangles ( CPST ) Corresponding Part of Similar Corresponding parts of congruent triangles are congruent" (CPCTC) is a succinct statement of a theorem regarding congruent trigonometry, defined as triangles either of which is an isometry of the other. |
|
| 11179. |
Sin/1-cos + tan/1+cos=cosecsec+cot |
| Answer» sinA/(1-cosA)+tanA/(1+cosA)=sinA[1/(1-cosA)+1/{cosA(1+cosA)}]=sinA[{(1+cosA)cosA+(1-cosA)}/{cosA(1+cosA)(1-cosA)}]=sinA(cosA+cos²A+1-cosA)/{cosA(1-cos²A)}=sinA(1+cos²A)/cosAsin²A=(1+cos²A)/cosAsinA=1/cosAsinA+cos²A/cosAsinA=secAcosecA+cotA | |
| 11180. |
if tan theta is equal to 2A and cot theta is to (A-18) find the value of A |
| Answer» 36 | |
| 11181. |
What is the solution for sin teta plus cos teta |
|
Answer» 1 1 |
|
| 11182. |
SinA +cosA=4 find the value of sinA_cosA |
| Answer» | |
| 11183. |
Euclid division method |
| Answer» a is equal to bq + r.whwer a os dividend , b is divisor , q is quotient and r is remainder. | |
| 11184. |
√x + y = 7√y + x =11 Find x and y |
| Answer» {tex}√x + y = 7 and √y + x =11{/tex}Now,\xa0{tex}√x=7-y{/tex}or,\xa0{tex}x=(7-y)^2{/tex}Putting the value into other equation,we get,{tex}√y+(7-y)^2=11{/tex}or,\xa0{tex}√y+49-14y+y^2=11{/tex}or,\xa0{tex}y^2-14y+√y+38=0{/tex}Let{tex}√y=m => y=m^2{/tex}Now,\xa0{tex}m^4-14m^2+m+38=0{/tex}or,\xa0{tex}(m-2)(m^3+2m^2-10m-19)=0{/tex}Therefore,{tex}m-2=0{/tex}or,\xa0{tex}m=2{/tex}As we know that,\xa0{tex}\\sqrt{y} =m{/tex}Then,\xa0{tex}√y=2{/tex}or,\xa0{tex}y=4{/tex}Also,\xa0{tex}√x=7-y{/tex}\xa0or,\xa0{tex}√x=7-4=3{/tex}or,\xa0{tex}x=9{/tex}\xa0\xa0 | |
| 11185. |
what are rational numbers |
| Answer» The numbers which can be written in the form of p/q are called rational numbers. | |
| 11186. |
what are rational numbers? |
|
Answer» The number that can be written in the form of p/q , where q is not equal to zero . Example = 2 , because it can be written in the form of p/q i.e 2/1 The number in the form of p/q where p and q are integers and q is not equal to zero . |
|
| 11187. |
Is herons formula correct78!#£#¥)) |
| Answer» Yes herons formula is correct | |
| 11188. |
How to spilliting equation class 10 |
| Answer» | |
| 11189. |
Formula of introduction of trigonatry |
| Answer» PBP/HHB | |
| 11190. |
the sum of the number and its reciprocal is 5/2.find the number. |
| Answer» Let the number be x and y.Now, (1/x)+(1/y)=5/2and x+y=5/2Therefore, x+y= (1/x)+(1/y)\xa0or, x+y=(x+y)/xyor, xy=1So, x = 2 and y = 1/2\xa0 | |
| 11191. |
Prove root 4 be iraratiinal |
|
Answer» It is not possible because 4 is a square of 2 . It would not be irrational because it is square root of,2 Prove {tex} \\sqrt{4}{/tex} is irrational number{tex} \\sqrt{4}{/tex} is = {tex} \\sqrt{2 x 2 }{/tex} = 2 which is a rationalHence {tex} \\sqrt{4}{/tex} is a rational number and cannot be proved irrational\xa0 Root 4 cannot be irrational as it is the square of 2 No |
|
| 11192. |
Prove that underroot 5is a irrational |
| Answer» let us assume to the contrary that root 5 is rational then it must in the form of p/q[q is not equal to 0][p and q are co-prime]root 5=p/q=> root 5 * q = psquaring on both sides=> 5*q*q = p*p ------> 1p*p is divisible by 5p is divisible by 5p = 5c [c is a positive integer] [squaring on both sides ]p*p = 25c*c --------- > 2sub p*p in 15*q*q = 25*c*cq*q = 5*c*c=> q is divisble by 5thus q and p have a common factor 5there is a contradictionas our assumsion p &q are co prime but it has a common factorso √5 is an irrational | |
| 11193. |
Mid term splitting - 2x²+5x-20 |
| Answer» | |
| 11194. |
Linear equation in two variables test paper |
| Answer» Let us take equations\xa02x - 4y = 0 2x - 2y = 2First try to simplify thase equations if possiblex - 2y = 0 .... i x - y = 1......... iix =2y ......... iii\xa0put the value of x in ii2y - y = 1y = 1put the value of y in iiix = 2 x 1= 2x = 2 , y = 1\xa0 | |
| 11195. |
if so is 2,4,6......find its nth rearm? |
| Answer» | |
| 11196. |
Hcf of (1001,910) |
| Answer» {tex}1001=13×7×11{/tex}{tex}910= 13×2×5×7{/tex}H.C.F is 13×7=91 | |
| 11197. |
Find all the zeros of the polynomial x^4+x^3-34x^2-4x+120, if two of its zeros are 2 and -2. |
| Answer» Given polynomial is x4 + x3 - 34x2 - 4x + 120Since, the two zeroes of the polynomial given is 2 and -2So, factors are (x + 2)(x - 2) = x2 + 2x - 2x - 4 = x2 - 4dividend = divisor × quotient + remainderdividend= x4 + x3 - 34x2 - 4x + 120divisor = x2 - 4quotient = x2 + x - 30remainder = 0So, x4 + x3 - 34x2 - 4x + 120 = (x2 - 4)(x2 + x - 30)= (x - 2)(x + 2)(x2 + 6x - 5x - 30)= (x - 2)(x + 2)(x + 6)(x - 5)Therefore, the zeroes of the polynomial = x = 2, -2, -6, 5 | |
| 11198. |
x+y/xy=2 |
| Answer» | |
| 11199. |
Find the area of rhombus of its vertices are A(3,0) B(4,5) C(-1,4) D(-2,-1) |
| Answer» {tex}\\sqrt{}{/tex}{3 - (-1)}2 + (0 - 4)2 + {4 - (-2)}2 +{\xa05 - (-1)}2{tex} \\sqrt{}{/tex}(3 + 1)2 + 16 + 36 + 36{tex}\\sqrt{}{/tex}16 + 16 + 36 + 36{tex} \\sqrt{}{/tex}104{tex} \\sqrt{}{/tex}2x2x262\xa0{tex} \\sqrt{}{/tex}26 | |
| 11200. |
Prove Pythagoras theorm |
| Answer» BC^2=AC^2-AB^2=(5)^2-(4)=25-16=9BC^2=(3)^2 | |