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.
| 22351. |
Prove that:: sinA/cotA+cosecA= 2+(sinA/ cotA-cosecA) |
| Answer» | |
| 22352. |
Can u give a link to download maths ncert pdf |
| Answer» Check in google | |
| 22353. |
Prove that the diagonals of a rectangle bisect each other and are equal using co-ordinate geometry |
|
Answer» Usme koi problem nahi hain coordinates position hote hain woh same nhi ho sakta But coordinate of vertices are not equal Find the midpoints of both the diagonals both will be sAme means |
|
| 22354. |
Determine if the following are in proprotion |
| Answer» | |
| 22355. |
Sin 4 theta + cos 4 theta |
| Answer» | |
| 22356. |
1+10+x+15=1x+1 |
|
Answer» 25 So what.... 25... |
|
| 22357. |
What is its syllabus |
|
Answer» All the 15 chapters r coming.... What did you mean ? Plz mention the subject ....☺ |
|
| 22358. |
Find the area of the sector of a circle whose radius is r and length of the arc is l |
|
Answer» 1/2*r*l sq.unit 1/2× r×l |
|
| 22359. |
Find the 20th term from the last term of the AP 3,8,13,......253 |
|
Answer» a child puts one 5 rupee coin of her saving in the piggy bank on the first day, 253=3+(n-1)×5n-1=250/5=50So n=51So 20th term from last is 51-20+1=32nd terma32=3+31×5=158 and \xa0a = 3, d = 5Now, 253 = a + (n + 1) d⇒ 253 = 3 + (n -1) x 5⇒ 253 = 3 + 5n – 5⇒ 5n = 253 + 2 = 255⇒ n = 255/5 = 51Therefore, 20th term from the last term = 51 – 19 = 32a32\xa0= a + 31d\xa0= 3 + 31 x 5= 3 + 155= 158Thus, required term is 158 a20=a+19d=3+19×5=98 The 20th term is 158 . 158 158 |
|
| 22360. |
Ex 1.1 question no. 4 |
|
Answer» Apply Euclids division algorithm. See example 2 then u will make it easy to solve this . Check out this app go to mathematics section then close click to ncert solutions u will get the answer...... |
|
| 22361. |
Ncert chapter 1 EX 1.1Question no. 3 |
| Answer» It\'s very simple to solve find the HCF of 616 , 32 then u got ur ans. 8 coloums. | |
| 22362. |
All the questions come from ncert book in this board rxam |
|
Answer» Mostly questions come from ncert but still you have to practice from other books too.. 80% from ncert and 20% from our acknowledge . From total marks 80 . It is not necessary but u have to understand the concept of each chapter then u are able to attempt any kind of question some question can also come from another extra books but don\'t take tension if your concept is clear u can cross any hurdle.....? |
|
| 22363. |
How many terms of AP -6,-11/2,-5......will give the sum zeroPls amswer fast |
|
Answer» Thank you 0=n/2(-12+(n-1)×(1/2))n(-12+ n/2 -1/2)=0Or -24+n-1=0n=25 |
|
| 22364. |
SinO-cosO+1÷ SinO+cosO-1 = 1 ÷ secO-tanO |
| Answer» It need to be evaluated | |
| 22365. |
Who invented geometry |
| Answer» euclid | |
| 22366. |
Find the HCF of 1190and 1445 in the. form of 1190m+1445n |
| Answer» Here we have to find HCF of 1190 and 1445 and express the HCF in the form 1190m + 1445n.1445 = 1190 ×\xa01 + 2551190 = 255 ×\xa04 + 170255 = 170 ×\xa01 + 85170 = 85 ×\xa02 + 0So, now the remainder is 0, then HCF is 85Now,85 = 255 - 170= (1445 - 1190) - (1190 - 255 ×\xa04)= 1445 - 1190 - 1190 + 255 ×\xa04= 1445 - 1190 ×\xa02 + (1445 - 1190) ×\xa04= 1445 - 1190 ×\xa02 + 1445 × 4 - 1190 ×\xa04= 1445 ×\xa05 - 1190 ×\xa06= 1190 ×\xa0(- 6) + 1445 ×\xa05= 1190m + 1445n , where m = - 6 and n = 5 | |
| 22367. |
Model test paper 1 math solutions |
| Answer» | |
| 22368. |
Find the value of cotsquaretheta +1/sinsquare theta |
| Answer» The value is 1/sin^4theta | |
| 22369. |
state pythagorus thorem |
| Answer» In a right angled triangle, the square of hypotenuse is equal to the sum of squares of other two sides. | |
| 22370. |
Can anyone send solutions to math QP code RSPL /1 |
| Answer» no | |
| 22371. |
If sec theta + tan theta =p then find the value of sin theta. |
| Answer» P square - 1 / p square + q | |
| 22372. |
1aur1 kitna hota hai |
|
Answer» Is questio ka jawab to computer bhi nahin de payega 3 Full syllabus complete kr liya kya.. Jo itne complicated question puch rahe ho?? 2 , 11 It\'s a very complicated question ????? |
|
| 22373. |
A pair of tangents can be constructed to a circle inclined at an angle of 170°.justify. |
|
Answer» Tangents are possible as angle between radius=360-(90+90+170)=10°Hence tangents are possible Take photo of figure then copy and paste.if there is problem try in Mozilla fire fox browser. Yes, it is possible but it seems to be very nearest to the circumference. How should we post the figure |
|
| 22374. |
Prove that sin-cos+1 1 __________ = ______. Sin +cos-1 Sec-tan |
|
Answer» Hope yu can understand ?? First divide numerator and denominator by cos theta, then multiply (tan theta - sec theta) in both numerator and denominator. And solve it. LHS: Sin-cos+1/sin+cos-1. = ((sin-cos+1)/cos)/(sin+cos-1)/cos=Tan+sec-1/tan-sec+1=== substitute sec^2-tan^2 in place of +1=Tan+sec-1/tan -sec+(sec^2-tan^2)=Tan+sec-1/tan-sec+(sec-tan)(sec+tan)=Tan+sec-1/-(tan-sec)(sec+tan-1)=1/sec-tan Explain the every thing Can\'t understand the question.? Sin -cos+1÷sin +cos -1 = 1÷sec-tan |
|
| 22375. |
7sin^2A+3cos^2A=4.S.T: tanA=1/√3 |
| Answer» 4sinAsq+3sinAsq+3cosAsq=43(sinAsq +cosAsq)+4sinAsq=43+4sinAsq=44sinAsq=1SinAsq=1/4SinA=√1/4=1/2SinA=30TanA=30TanA=1/√3 | |
| 22376. |
If the equation(1+m2)x2+2mcx+(c2-a2)=0 has equal roots ,prove that c2=a2(1+m2) |
|
Answer» Given- (1+m2)x2+2mcx+(c2-a2)=0If this equation has equal roots then B2-4ac = 0(2mc)2-4(1+m2)(c2-a2)=04m2c2-4(1+m2)(c2-a2)=04m2c2 = 4(1+m2)(c2-a2)m2c2 = (1+m2)(c2-a2)m2c2 = c2 - a2 + m2c2 - m2a2 0 = c2 -a2 - m2a2 m2a2 = c2-a2 m2a2 + a2 = c2 (1+m2)a2 = c2 HENCE PROVED ???? Not knowing |
|
| 22377. |
Formula of (a-b) whole cube |
|
Answer» a3-b3-3ab (a-b) Sorry it is ( a cube -b cube-3,ab(a-b) (a cube -b cube)(a sq. +ab+b sq.a) |
|
| 22378. |
How to prove irrational |
|
Answer» Let me tell u the format first with the help of an example I am taking √5 here:Let us assume that\xa0√5 is a rational number.we know that the rational numbers are in the form of p/q form where p,q are intezers.so,\xa0√5 = p/q p =\xa0√5qwe know that \'p\' is a rational number. so\xa0√5 q must be rational since it equals to pbut it doesnt occurs with\xa0√5 since its not an intezertherefore, p =/=\xa0√5qthis contradicts the fact that\xa0√5 is an irrational numberhence our assumption is wrong and\xa0√5 is an irrational number.hope it helped u ☺️ Firstly u need to assume that the given number is rational then u have to express it in the form of p/q. At last u will get a wrong equation which contradicts our assumption. |
|
| 22379. |
How can we find slope in coordinate geometry |
| Answer» | |
| 22380. |
If nth term of an AP is 2n+1.then what is the sum of its first three terms? |
|
Answer» An = 2n + 1Let n be 1a1 = 2(1) + 1a1 = 3Let n be 2a2 = 2(2) + 1= 4 + 1= 5Let n be 3a3 = 2(3) + 1a3 = 7a1 + a2 + a3 = 3 + 5 + 7 = 15\xa0 First put n=1and then put n=2 u will get the 1st term and the 2nd term and then find d 15 |
|
| 22381. |
The larger of two supplementary angles exceeds the smaller by 20 degrees. Find them. |
|
Answer» Thanks let measure of smaller supplementary angle = x°Larger angle = x + 20°As per given conditionx + x + 20 = 180°2x = 180° - 20°2x = 160°x = 160/2x = 80°So, angles are 80° and 100°. 80°and100° |
|
| 22382. |
SecA -TanA=P find the value of sinA |
| Answer» 1-psquare divided by 1 +p square | |
| 22383. |
Tan A/1-cotA=cotA/1-tanA=1+tanA+cotA prove |
| Answer» It is not possible keep a=45 degrees | |
| 22384. |
an+d= |
| Answer» an=a+(n-1)dan=a+nd-dan+d=a+nd | |
| 22385. |
Hii guys how is your going for boards |
|
Answer» Thik hi Aap apna batao How is your practice going for boards |
|
| 22386. |
How many numbers I will got in my boards examination after downloading and studying this app???... |
|
Answer» Wo to aap ki study pe depend karta h Pata Nahi Nisha ji ? 101 |
|
| 22387. |
How can we find median? |
|
Answer» There are 2 methods1. By impirical formula that is 3med. -2mean=mode2.by median formula l+(n/2-cf)÷f×h By applying formulae |
|
| 22388. |
√(-a cos alpha-a sin alpha)^2 + (b sin alpha-(-b cos alpha))^2 |
|
Answer» Or, a^2 + b^2 + 2SinaCosa( a^2 - b^2)But both are same (a^2 + b^) 1 - 4sin^2a cos^2aWhere a is alpha |
|
| 22389. |
In an Ap the first nth term is -17 and the sum of n terms is 60 .find the common difference |
| Answer» Gravitational field intensity = gradient of Gravitational potential = dV/drThe gravitational potential gradient dV/dr is defined as the rate of change of gravitational potential with distance in the field. This is equal to the gravitational field intensity at that point.U = gravitational potential energy = –\xa0GMm/rGravitational potential V = U/m = – GM/rGravitational field intensity I = dV/dr= GM/r2 | |
| 22390. |
√5 ek prem sankhya hai |
|
Answer» What is meant by prem sankhya? Ur ryt What is Prem sankhya define or do not write such senseless.this not your FB. What ???? |
|
| 22391. |
For some a and b, if HCF of 55 and 210 is 210a + 55b, then find value of a and b |
| Answer» Value of a is 5 and b is - 19 ......Hope it helps u ? | |
| 22392. |
1 ball 2 bat so |
|
Answer» Plz explain ur question a bit more .. What\'s the question? |
|
| 22393. |
Chapterwise blue print of maths pls share if anyone havr |
|
Answer» You can get it from cbse official website I have the blue print but how can I share here? |
|
| 22394. |
Class 10 board exam 2019 ka project work kaun 2 banana h please help me |
| Answer» | |
| 22395. |
Cosec²63° + tan²24°/cot ²66° + sec²27° |
|
Answer» 2 2 2 |
|
| 22396. |
235+456 |
|
Answer» Hamare to 17 se h 691 691 No.. 691 691......0 781 O...... |
|
| 22397. |
Find out the HCF and LCM of 24, 15 and 36 by prime factorization. |
|
Answer» 24 = 2 × 2 × 2 × 336 = 2 × 2 × 3 × 315 = 3 × 5HCF = 3\xa0LCM = 2 × 2 × 2 × 3 × 3 × 5 = 360 HCF and LCM are 3&360 respectively |
|
| 22398. |
We write tan =sin / cos...... And cos = 1/sec..so why we can\'t write cos =sin/tan........ |
|
Answer» Dedia.... Most welcome.. Ab aap BHI mere question ka motivational answer de dijiye Thanks for your motivational answer..... ? Because tan is equal to sin/cos is an identity which we can use directly but cos is equal to sin/tan is derived from that. |
|
| 22399. |
Sin(A+B) =sinA+sinB |
| Answer» No, they r not equal | |
| 22400. |
If p ,q and r are in AP , then prove that (p+2q-r)(2q+r-p)(r+p-q)=4pqr |
| Answer» | |