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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.
| 22651. |
Pls tell chapter vise waitage for board exam of class 10 maths |
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Answer» You can see by this app only Number system - 06, Algebra - 20, Coordinate geometry - 06, Geometry - 15, Trigonometry - 12, Mensuration - 10, Statistics and probability - 11 hope it will help you ☺☺☺☺☺ |
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| 22652. |
If sec theta +tan theta=p then find value of cosec theta. |
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Answer» Hey I got the answer it can be solved by sec square - tan square theta Hi , sorry I am not getting your point. Hi P square +1/p square -1 and -1 |
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| 22653. |
4/x –3=5/2x+3 |
| Answer» 4/x-3=5/2x+34/x-5/2x=68x-5x=12x²3x=12x²3=12x²/x3=12xX=3/12X=1/4 | |
| 22654. |
1x+2y |
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Answer» I agree with rashi Right Jazzy aap plz ek baar check kar lo ki question thik hai ya nhi,may be question incomplete hai. It has no answer because in this equation two variable |
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| 22655. |
Use Euclid division leema to show that cube of any positive integer is of the from 9m,9m+1or9m+8 |
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Answer» According to Euclid Division Lemma,a=bq+r,b Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.So, we have the following cases :Case I : When x = 3q.then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.Case II : When x = 3q + 1then, x3 = (3q + 1)3= 27q3 + 27q2 + 9q + 1= 9 q (3q2 + 3q + 1) + 1= 9m + 1, where m = q (3q2 + 3q + 1)Case III. When x = 3q + 2then, x3 = (3q + 2)3= 27 q3 + 54q2 + 36q + 8= 9q (3q2 + 6q + 4) + 8= 9 m + 8, where m = q (3q2 + 6q + 4)Hence, x3 is either of the form 9 m or 9 m + 1 or, 9 m + 8. Bshshshghdh |
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| 22656. |
1-sin01+sin0 |
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Answer» Ya the answer is correct. ( cos square theta). Hi =(1-sinΦ)(1+sinΦ)=1-sin²Φ=Cos²Φ What\'s the question bro I can\'t understand?? |
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| 22657. |
Perimeter of trangle |
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Answer» Sum of all sides of triangle Perimeter of polygon = addition of all sidesPerimeter of triangle = Adding all 3 sidesPerimeter of equilateral triangle= 3 × side AB + BC + CA (add all three sides of ∆) A+b+c |
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| 22658. |
If 1 polynomial is divided by other polynomial and get remainder Give me a example with answer |
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| 22659. |
If px squar +3x+q=0 has two roots x=-1,x=-2,find p-q |
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Answer» Nice px2 + 3x + q = 0x = 1p + 3 + q = 0p + q = -3 .......... (i)x = -2p(-2)2 + 3(-2) + q = 04p - 6 + q = 04p + q = 6 ........... (ii)Subtract (i) and (ii)3p = 9p = 9/3 = 3Put it in (i)3 + q = -3q = -3 -3q = -6 |
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| 22660. |
Area of the minor segment and major segment |
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Answer» The segment of a circle divides it into two region namely major segment and minor segment. The segment having larger area is known as the major segment and the segment having smaller area is known as minor segment.Area of minor segment = θ/360 πr2 |
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| 22661. |
8.2 |
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Answer» Which question Aapka question kya hai Bhai ? |
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| 22662. |
If the point A(a,2) is equidistant from the points B(8,-2) andC(2,-2),find the value of a |
| Answer» we have that:AB = ACsoAB² = AC²then:(a-8)² + (2 + 2)² = (a-2)² + (2+2)²a²- 16a + 64 +16 = a²- 4a + 4 +16-16a + 64 = -4a + 4-16a + 4a =4 - 64-12a = -60;12a = 60;a = 60/12 = 5;a = 5The value of a is 5.\xa0 | |
| 22663. |
Cos - sin =✓sin prove that Cos + sin =✓sin |
| Answer» Ur question may be wrong | |
| 22664. |
If a/b=b/c then the value of 1/b-c + 1/b-a is ____ |
| Answer» | |
| 22665. |
How many ordered pairs of (x,y) integers satisfy x/15=36/y? |
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| 22666. |
find the value of r |
| Answer» Incomplete question... | |
| 22667. |
Prove that the points (3,0),(6,4),(-1,3)are the vertices of a right angled isosceles triangle |
| Answer» Given A(3, 0), B(6, 4) and C(-1, 3)AB2\xa0= (3 - 6)2\xa0+ (0 - 4)2\xa0= 9 + 16 = 25BC2\xa0= (6 + 1)2\xa0+ (4 - 3)2\xa0= 49 + 1 = 50CA2\xa0= (-1 - 3)2\xa0+ (3 - 0)2\xa0= 16 + 9 = 25AB2\xa0= CA2\xa0or, AB = CATriangle is isocelesAlso, 25 + 25 = 50or, AB2\xa0+ CA2\xa0= BC2Since, pythagoras theorem is verified, therefore triangle is right-angled triangle. | |
| 22668. |
If tan A= ntan B and sin A= msinB , then prove that cos squre A = m square -1 / n square -1 |
| Answer» Given,\xa0tan A = n tan B{tex} \\Rightarrow{/tex} tanB = {tex}\\frac{1}{n}{/tex}tan A{tex}\\Rightarrow{/tex}\xa0cotB =\xa0{tex}\\frac { n } { \\tan A }{/tex}..........(1)Also given,\xa0sin A = m sin B{tex}\\Rightarrow{/tex}\xa0sin B =\xa0{tex}\\frac{1}{m}{/tex}sin A{tex}\\Rightarrow{/tex}\xa0cosec B =\xa0{tex}\\frac { m } { \\sin A }{/tex}.....(2)We know that, cosec2B - cot2B = 1, hence from (1) & (2) :-{tex} \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } } { \\tan ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = sin2A{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = 1 - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = n2cos2A - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = (n2 - 1) cos2A{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - 1 } { n ^ { 2 } - 1 } ={/tex}\xa0cos2A | |
| 22669. |
9609 root |
| Answer» 98.02 | |
| 22670. |
draw a triangle ABCwith sides 6cm ,8cm,9cm |
| Answer» First you draw a line of 9cm then take angle on scale of 6 and 8 cm mark an arc then draw a line | |
| 22671. |
The length of the shadow of a person is √3 times of its hight. Find the angle of elevation of sun |
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Answer» 30° is angle of elevation of sun ⛅ 30° Let the height of the person be hTherefore, the height of the shadow is /3 h.Tan=h÷/3hTan=1÷/3Tan30=1÷/3Therefore angle of elevation is 30° |
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| 22672. |
Gdj |
| Answer» Hi | |
| 22673. |
The perimeter of a sector of a circle of radius 8 cm is 25 m find the area of sector |
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Answer» Sry....this solution is wrong If perimeter and radius is given,we can find the length of the arc.25=8+8+length of arcLength of the arc =25-16=9cmNow we can find the angle of the sector 9cm= angle÷360×2×22÷7×8Angle=5670÷88Area of the sector=angle÷360×22÷7×8×8 =5670×22×8×8÷88×360×7 =36cmsq. |
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| 22674. |
CosecA+1= cosACosecA-1 1-sinA |
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| 22675. |
Tignomentri me kitne formule h |
| Answer» अनगिनत | |
| 22676. |
If sinA=1/3 then find the value of 2 cot2A +2 |
| Answer» 18.... | |
| 22677. |
Sin2/sin7 |
| Answer» Sin of small angle is same as the angles so here the value is 2/7 | |
| 22678. |
Chapter 14 in maths question important |
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Answer» Example 8 this type of questions Will come and exercise 14.3 question 2 and exercise 14.3 is very important one question of 5 marks will come always In NCRT book |
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| 22679. |
Cos +tan/sin+cos 1 |
| Answer» What is the 1,^or° | |
| 22680. |
55.23 + 23.56 |
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Answer» 78.79 78.79 78.79 |
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| 22681. |
draw a line segment of length 7.6 cm and divide it in the ratio 5:8 measure the parts |
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Answer» Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.triangle whose sides are times the corres ponding sides of the isosceles triangle. answer? |
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| 22682. |
X=p sec+q tan and y=p tan +q sec prove x^2-y^2=p^2-q^2 |
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| 22683. |
Find points at which ,pair of equations x=a and y=b intersect, when graphically represent |
| Answer» intersecting at (a, b) | |
| 22684. |
How score 100% maths. |
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Answer» Learn all formula and lots of practice again and again....... Keeps practice........ Do pratice again and again Just keep studying ,u will score best Work out well |
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| 22685. |
Sin theta equal to cos theta then find the value of theta |
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Answer» Usme baat krege Apna ek ques. Send kr Kumkum Hii Hlo shivangi & Sushma Thete = 45° 45 |
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| 22686. |
State the fundamental theorem of arithematic |
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Answer» Thanks karuna Every composite number can be expressed as a product of primes ,and this factorisation is unique,apart from the order in which the prime factors occur |
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| 22687. |
If cosec theta + Cot theta = x , find the value of cosec theta - Cot theta |
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Answer» By identity cosec2a - cot2a =1Using identity a2-b2=[a+b][a-b] we have(Coseca+cota) (coseca- cota)=1 (by identiy at top)= x(coseca- cota)= 1( given =x)(coseca-cota)= 1/x ANSWER!! Cosec²Q-Cot²Q= 1...(CosecQ-CotQ) x (Cosec+CotQ) = 1....(CosecQ-CotQ) X (x) = 1.....CosecQ-CotQ = 1/x.. |
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| 22688. |
2x+ |
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| 22689. |
Find the distance between the points (a cos thita , a sin thita ) and ( - a cos thita , a sin thita) |
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Answer» How u do this sushma ca u explain 2cos thita Please answer me |
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| 22690. |
Find the value of cos([email\xa0protected])[email\xa0protected]÷[email\xa0protected] |
| Answer» | |
| 22691. |
Whai is polynomial |
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Answer» It is a degree of constant Polynomial is a degrre of constant |
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| 22692. |
proof ^5.is.iresenal |
| Answer» It is √5 is irrational prove that | |
| 22693. |
Find the value of p for which one of quadratic equation pxka square - 14x+a =0 is 6 times the other |
| Answer» Let\xa0{tex}\\alpha{/tex}\xa0and\xa0{tex}6\\alpha{/tex}\xa0be the roots of equation.We have, {tex}px^2-14x+8=0{/tex} where a= p, b = -14, c = 8Sum of zeroes{tex} = -\\frac ba = -\\frac{-14}{p}{/tex}{tex}\\alpha +6\\alpha=\\frac{14}{p}{/tex}{tex}7\\alpha = \\frac{14}{p}{/tex}{tex}\\alpha = \\frac2p{/tex}............(i)Also, Product of the zeroes\xa0{tex} = \\frac 8p=\\frac ca{/tex}{tex}\\alpha \\times 6\\alpha = \\frac 8p{/tex}{tex}6\\alpha^2=\\frac 8p{/tex}From (i){tex}6(\\frac{2}{p})^2=\\frac 8p{/tex}{tex}6\\times \\frac {4}{p^2}=\\frac 8p{/tex}{tex}\\frac{6}{p^2}=\\frac2p{/tex}{tex}\\frac 62=\\frac{p^2}{p}{/tex}{tex}Hence, \\ p=3{/tex} | |
| 22694. |
Derivation |
| Answer» Of what | |
| 22695. |
Prove of theorem 6.3 of ch6 in ncert book |
| Answer» Ye to book me hai hi | |
| 22696. |
Ogive method from stats Ch. |
| Answer» Means??? | |
| 22697. |
Tan1( |
| Answer» Pls tell me what the method of ogive from stats Ch. | |
| 22698. |
cbse deleted lessons |
| Answer» In mathematics all the 15 chapters are coming none of the chapter is deleted. | |
| 22699. |
(a+b) (a-b) = |
| Answer» a2-b2 | |
| 22700. |
Can the 6n ,n being a natural number ,end with the digit 5 ? Give reason |
| Answer» No , as any no. ending with 5 have its multiples An odd number and 5 . Here 6 is even so it is not possible | |