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.
| 6651. |
If p and q are the zeroes of the equation x^2 +px-q ..Then find the value of p and q |
| Answer» | |
| 6652. |
Find the median of the following dataMarks 20-30 30-40 40-5050-6060-7070-80No.of students51525207810 |
| Answer» | |
| 6653. |
Simplify cosec2 A÷ 1+ cot2 A |
| Answer» If 2 is square of the ratio then the answer is 1 ans:- cosec 2A =1÷sin2A and. 1+cot2A = 1 + cos2A ÷sin2A. 1÷sin2A÷sin2A +cos2A ÷sin 2A. Here , sin 2A get cut then. Remained is 1÷sin2A + cos2A. ( Sin2A +cos2A=1) 1÷1= 1 | |
| 6654. |
If x=a cos(theta), y=b sin(theta) then find the value of b^2x^2+a^2y^2-a^2b^2. |
| Answer» | |
| 6655. |
Tan 3x = 1/ tan 2x , solve |
| Answer» | |
| 6656. |
∆abc~∆def and their perimeters 32cm and 24cm respectively. If ab=10cm, find de |
| Answer» De = 7.5 | |
| 6657. |
Cosec A (secA – 1)– cot A (1–cos A)= tanA – sinA |
| Answer» | |
| 6658. |
Prove that √2 is a irrational number |
| Answer» | |
| 6659. |
Prove that √2 is an irrational number |
| Answer» | |
| 6660. |
Gddhikvdik |
| Answer» | |
| 6661. |
(234+453)(234-567)/2 |
| Answer» -114385.5 | |
| 6662. |
(987654321×1234567890)+(45632÷324)-(35432+65764)/4364 |
| Answer» isse solve krne se ashaa hum apna maths ka homework krleee | |
| 6663. |
Prove irrational √n+1+√n-1 |
| Answer» +1&-1 willcut out each other.Then, √n+√n will become2√n .√n is irratinal no.& if any irrational no. Multiplies with any rational no. The product is irrational. So √n+1+√n-1 is irrational. | |
| 6664. |
PS is the bisecter of triangle PQR.prove that QS÷SR=PQ÷PR |
| Answer» Given: In figure. PS is the bisector of\xa0{tex}\\angle QPR{/tex} of\xa0{tex}\\triangle PQR{/tex}To prove :{tex}\\frac { Q S } { S R } = \\frac { P Q } { P R }{/tex}Construction: Draw RT || SP to meet QP produced in T.Proof:\xa0{tex}\\because {/tex}\xa0RT||SP and transversal PR intersects them{tex}\\therefore \\angle 1 = \\angle 2{/tex}\xa0(1) ..... Alt. Int.\xa0{tex}\\angle s{/tex}{tex}\\because {/tex}\xa0RT||SP and transversal QT intersects them{tex}\\therefore \\angle 3 = \\angle 4{/tex} (2), ..... corres.\xa0{tex}\\angle s{/tex}But\xa0{tex}\\angle 1 = \\angle 3{/tex}\xa0...... Given{tex}\\therefore \\angle 2 = \\angle 4{/tex}\xa0......From (1) and (2){tex}\\therefore P T = P R{/tex}\xa0(3) ......{tex}\\because {/tex}\xa0Sides opposite to equal angles of a triangle are equal\xa0Now in\xa0{tex}\\Delta \\mathrm { QRT }{/tex}PS = RT .......By construction{tex}\\therefore \\frac { Q S } { S R } = \\frac { P Q } { P T }{/tex}\xa0....... By basic proportionally theorem\xa0{tex}\\Rightarrow \\frac { Q S } { S R } = \\frac { P Q } { P R }{/tex}\xa0......From (3) | |
| 6665. |
Is 0 a even number? If yes, then how and why? |
| Answer» Zero is an even number. In other words, its parity—the quality of an integer being even or odd—is even. The simplest way to prove that zero is even is to check that it fits the definition of "even": it is an integer multiple of 2, specifically 0 × 2. | |
| 6666. |
NCERT class x .ch triangles ,ex- 6.3, ques no.- 10 & 14. |
| Answer» | |
| 6667. |
7+77+777+............+777.......7=7/81(10n+1 - 9n -10) |
| Answer» Sn = 7 + 77 + 777 + ---- + n terms{tex}= \\frac { 7 } { 9 } [ 9 + 99 + 999 + - - - - + n \\text { terms } ]{/tex}{tex}= \\frac { 7 } { 9 }{/tex}[(10 - 1 ) + (102\xa0- 1) + (103\xa0- 1) + --- + n terms]{tex}= \\frac { 7 } { 9 }{/tex}[(101 + 102 + ---- +n terms) - (1 + 1 + 1 ---- + n terms)]{tex}= \\frac { 7 } { 9 } \\left[ \\frac { 10 \\left( 10 ^ { n } - 1 \\right) } { 10 - 1 } - n \\right]{/tex}{tex}= \\frac { 7 } { 9 } \\left[ \\frac { 10 \\left( 10 ^ { n } - 1 \\right) } { 9 } - n \\right]{/tex} | |
| 6668. |
Sin2 A = 1/cosec2 A Is this true ?? |
|
Answer» Can u tell me how ? Yes |
|
| 6669. |
31-20 |
| Answer» 11 | |
| 6670. |
2-3 |
| Answer» | |
| 6671. |
what are the roots of 12-75x |
| Answer» X=4/25. | |
| 6672. |
Please mathamatics book in hindi |
| Answer» | |
| 6673. |
Px=xsquare+7x+9is divided by gx ,we get (x+2)and -1 as quetient and remainder respectively .find gx |
| Answer» Quadratic equation, p(x) = x2 + 7x + 9Quotient, q(x) = x + 2Remainder, r(x) = - 1Divisor, g(x) = ?p(x) = g(x) q(x) + r(x)x2+ 7x + 9 = g(x) (x + 2) - 1x2+7x+9+1 = g(x) (x+2)\xa0x2+7x+10 = g(x) (x+2){tex}\\therefore{/tex}\xa0g(x) =\xa0{tex}\\frac { x ^ { 2 } + 7 x + 10 } { x + 2 }{/tex}g(x) =\xa0{tex}\\frac {x^2+2x+5x+10}{x+2}{/tex}g(x) =\xa0{tex}\\frac {x(x+2)+5(x+2)}{x+2}{/tex}{tex}= \\frac { ( x + 2 ) ( x + 5 ) } { ( x + 2 ) }{/tex}g(x) = x + 5 | |
| 6674. |
Rational no. Between 2and4 |
|
Answer» Infinite. 3 |
|
| 6675. |
Hvh |
| Answer» | |
| 6676. |
Walhat is real root |
| Answer» | |
| 6677. |
Root 5 is irrational prove |
| Answer» Lets assume root 5 is rational. Root 5 =a/b where a and b are co-prime b is not =0b root 5 =asquaring both sides 5b square =a square - - - - - - 15 divides a square =5 divides aa= 5ca square =25c square\t--put value of a square in 15b square =25c square (cutting both) b square =5c square 5 divides by b square=5 divides b and we know 5 divides a but a and b are co-prime = our supposition is wrong. = root 5 is irrational. | |
| 6678. |
Will this year question paper become easy ? |
|
Answer» No because in beginning everything is hard but not for all. So please study and do not lie on luck. yes . bcz they collect only principal points throughout the book |
|
| 6679. |
Find quadratic polynomial sum and pooduct 1,1 |
| Answer» x*-1x+1 | |
| 6680. |
What is consistent? |
| Answer» In case of linear equations in two variables this term is used....when there is unique solution or many solutions,the system of equations is said to be consistent. | |
| 6681. |
Solve for X : 2-1/2-1/2-1/2-x |
| Answer» I think........... 2-x | |
| 6682. |
Prove that 3+2=5 |
| Answer» Squring both the sides | |
| 6683. |
Exercise 8.3 example 14 |
| Answer» | |
| 6684. |
Evaluate cos1.cos2.cos3...........cos180 |
| Answer» Oh awesomeCos1.cos2........Cos90......cos180(Cos90=0)=0 | |
| 6685. |
2n is a negative no. |
| Answer» | |
| 6686. |
Use Euclid \'s division algorithm to find the HCF of 10224 and 9648. |
| Answer» Hoshiyara na dikhao ,book me hai. Chup chap kar lo? | |
| 6687. |
In which ratio the segment joining point (-3,-4),(1-2)Divided by Y-axis |
| Answer» | |
| 6688. |
x2+5x=1200 |
| Answer» | |
| 6689. |
Simple ways for applications of trignomatry |
| Answer» | |
| 6690. |
2_3 |
| Answer» | |
| 6691. |
P (x)=(k+4) x square+13x+3k |
| Answer» | |
| 6692. |
What is difference between rational number and irrational number? |
|
Answer» Koi nahi kai. simple cheeze kisi ko ma pats ho to poochh Lena hi sahi very good for rising questions Rational no. can be written in the form of a/b where b not= 0 if a number could not be written in this form then it is irrational Itna bhi na pta |
|
| 6693. |
Math question |
| Answer» | |
| 6694. |
Last year paper for math |
| Answer» | |
| 6695. |
x²+3x-p By quadratic equation |
| Answer» Find it please | |
| 6696. |
SinA+cosecA=2Then prove sin6A+ cosec6A=? |
| Answer» | |
| 6697. |
2356+1 |
|
Answer» 2357 Answer is 2357 2357 |
|
| 6698. |
Formula of cross multiplication |
| Answer» | |
| 6699. |
Polymer |
| Answer» | |
| 6700. |
The product of two consecutive positive integers is 306 we need to find the integers |
|
Answer» Let the 1st number be x . So the 2nd number be x+1. Acc. to question , (x)(x+1)=306 After solving this we get x^2 + x = 306. Then x^2 + x - 306 = 0 . Then by middle term factorization, we get x^2 + 17x - 18x - 306 = 0. Then x(x + 17) - 18(x + 17). Then (x - 18) (x - 17) = 0. Then x = 17 and x= 18. Therefore the integers are 17 and 18 17 and 18 X(x+1)=306X^2+x-306=0Simplyfy the qudratic equation and u will get the answer as 17 and 18 |
|