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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.
| 21901. |
tanA=ntanB and sinA=msinB,prove that cos square A=m square-1/n square-1 |
| Answer» | |
| 21902. |
If the points A(x,2)B(-3,-4)and C(7,-5) are collinear then find the value of x |
| Answer» By using the formula of area of triangle 1/2{(x1(y2-y3)+x2(y3-y1)+x3(y1-y2)}.Area of triangle=0{because points are collinear}.We get the value of x. | |
| 21903. |
For what value of p,x=b is a zero of polynomial x^2-(a+b)x+a(p-2b)? |
| Answer» P=3b | |
| 21904. |
Maths compartment solution lat year paper |
| Answer» | |
| 21905. |
Which term of Ap will be zero 24,21,18,15....? |
| Answer» a=24d=-3a(n) =00=a+(n-1)d0=24+(n-1)×-3-24/-3=(n-1)8+1=nn=9 | |
| 21906. |
Solve by substitution method 3x-y=3 9x-3y=9 |
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Answer» Given,3x - y =\xa03,.............(i)9x - 3y\xa0= 9............(ii) It has Infinite many Solutions |
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| 21907. |
Sum of n term of an AP is (5n^2+3n)/2.find the Ntn term. |
| Answer» Nth term is 1/4. But i m not sure it is rgt or wrong ... | |
| 21908. |
If α and β are zeroes of polynomial x²-x-1 than find a polynomial with zeroes (2α+β) and (2β+α) |
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Answer» Explain 1 |
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| 21909. |
Find the area of a triangle whose vertics are (1,-1)(-4,6)and(-3,-5) |
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Answer» Let the vertices of triangle be A(1, -1), B(-4, 6) and C(-3, -5).Then we havex1 = 1, y1 = -1x2 = -4, y2= 6and x3= -3, y3 = -5 24 square meters....... |
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| 21910. |
Prove thatSec^2Q+COSEC^2Q=SEC^2Q.COSEC^2Q |
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| 21911. |
-15 is rational |
| Answer» No rational | |
| 21912. |
What is the value of cos^2 67°- sin^2 23° |
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Answer» Sin^2(90-67)° -Sin^2 23°Sin^2 23°-Sin^2 23°0 Ans is 0As Cos²67°-cos²67° = 0[Sin(90-theta)=cos theta] |
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| 21913. |
If the root of the equa |
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Answer» Write the question properly.☝ Plz, check yur question....... I don\'t understand your question. |
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| 21914. |
Find acute angle thita if √3 sin thita = cod thita |
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Answer» 30° √3 sinΦ = cosΦ=> √3 = cosΦ/sinΦ=> cot30 = cotΦ (cot30 = √3)Therefore, Φ = 30 degree Theeta =30° |
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| 21915. |
CSA of cylinder is 264m2 and volume is 924m3 . Find the ratio of their height upon diameter. |
| Answer» Ratio of height to diameter = 3/7 | |
| 21916. |
X+√y=21 and √x+y=29 |
| Answer» | |
| 21917. |
Sample question paper summative assessment-ll class-x 2016-17 mathmatic |
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| 21918. |
If Sn denots the sum of first n terms of an AP , prove that ,S12=3(S8-S4). |
| Answer» Let a be the first term and d be the common difference of the given AP. Then,Sn= {tex}\\frac{n}{2}{/tex}{tex} \\cdot {/tex}[2a+(n-l)d],{tex}\\therefore{/tex}\xa0{tex}3(S_8-S_4) = 3{/tex}[{tex}\\frac{8}{2}{/tex}{tex}(2a+7d)-{/tex}{tex}\\frac{4}{2}{/tex}{tex}(2a+3d)]{/tex}= {tex}3[4(2a+7d)- 2(2a+3d)] = 6(2a+11d){/tex}{tex}= \\frac { 12 } { 2 } \\cdot ( 2 a + 11 d ) = S _ { 12 }{/tex}.Hence, S12= 3(S8-S4). | |
| 21919. |
Find the firs four term of an A.P whose first term is 3x-y & comman difference is x-y |
| Answer» {tex}a_1 = 3x + y {/tex}and {tex}d = x - y{/tex}{tex}a_2 = a_1 + d = 3x + y + x - y = 4x{/tex}{tex}a_3 = a_2 + d = 4x + x - y = 5x -y{/tex}{tex}a_4 = a_3 + d = 5x - y + x - y = 6x -2y{/tex}So, the four terms are {tex} 3x + y, 4x, 5x - y {/tex} and{tex} 6x - 2y{/tex}. | |
| 21920. |
Find the firs term of an A.P whose first term is 3x+y & comman difference is x-y. |
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Answer» The first four terms of an A.P is 3x + y , 4x, 5x - y & 6x - 2y By the way what\'s the ans i wanna know....may be mine is correct Perhaps.....3x+y...4x.....5x-y.....-2y..... Sorry it is first four terms What u want to ask first term or nth term |
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| 21921. |
Given that tan thita =1/root 5 |
| Answer» | |
| 21922. |
Why am i don\'t see maths sample paper 5 |
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Answer» Are you asking about that sample papers which are provided by this app then you must have to buy sample papers as you have bought earlier sample paper 2-4. What is ur querry |
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| 21923. |
Formula for csa of cone |
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Answer» πrl πrl Cone is a three-dimensional structure having a circular base where a set of\xa0line segments, connecting all of the points on the base to a common point called apex.Curved surface area of a cone = |
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| 21924. |
2+3=p/q is irrational |
| Answer» 2+3 is rational no. | |
| 21925. |
Write AAA similarity criteria |
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Answer» Angle Angle Angle - AAA R.d sharma ya to ncert ko refer karo |
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| 21926. |
If a,b and c are in AP ,then b-a/c-a is |
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Answer» But my answer is 1 Since b-a is d and c-a=b+d-ab-a/c-a = d/b-a+d= d/2d=1/2 |
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| 21927. |
Reall no. |
| Answer» Real no are the normal no like 1,2,3,4,5etc | |
| 21928. |
Triangle thik se nhi hota kai kre bolo na koi?? |
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Answer» Brain se What do u mean by this?? |
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| 21929. |
If the polynomial 3x keep |
| Answer» ????? Complete your question ....... | |
| 21930. |
Type of questions in board exam |
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| 21931. |
Prove that: (cosec p - sin p)(sec p- cos p)(tan p + cot p) = 1 |
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Answer» One upon sin p Yaar anjali tum trigonometry kaisa karte ho .you are a legend .please mujhe bhi sikha do . Taking LHS,(cosec p- sin p)(sec p- cos p)(tan p+cot p)=(1/sin p-sin p)(1/cosp-cos p)(sinp/cos p+cos p/sin p)=(1-sin^2p/sin p)(1-cos^2p/cos p)(sin^2p+cos^2p/sin p×cos p)(Cos^2p/sin p)(sin^2p/cosp)(1/sin p×cos p)=1×1×1=1=RHS=proved =(1/sinp-sinp)(1/cosp-cosp)(sinp/cosp+cosp/sinp)=(1-sin²p/sin)(1-cos²p/cos)(sin²p+cos²p/sinpcosp)=(Cos²p/sinp)(sin²p/cosp)(sin²p+cos²p/sinpcosp)= 1 |
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| 21932. |
Theorm is compulsury come from traingle circle |
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Answer» Yesssss Yess it comes for 4 marks usually..soo it is compulsory Ya ofcourse |
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| 21933. |
Proove that:-tan÷1-cot+cot÷1-tan=1+tan+cot |
| Answer» {tex}\\frac { \\tan \\theta } { 1 - \\cot \\theta } + \\frac { \\cot \\theta } { 1 - \\tan \\theta } = 1 + \\tan \\theta + \\cot \\theta{/tex}{tex}\\text { L.H.S. } = \\frac { \\tan \\theta } { 1 - \\cot \\theta } + \\frac { \\cot \\theta } { 1 - \\tan \\theta }{/tex}{tex}= \\frac { \\frac { \\sin \\theta } { \\cos \\theta } } { 1 - \\frac { \\cos \\theta } { \\sin \\theta } } + \\frac { \\frac { \\cos \\theta } { \\sin \\theta } } { 1 - \\frac { \\sin \\theta } { \\cos \\theta } }{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\cos \\theta ( \\sin \\theta - \\cos \\theta ) } - \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}= \\frac { \\sin ^ { 3 } \\theta - \\cos ^ { 3 } \\theta } { \\sin \\theta \\cos \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}= \\frac { ( \\sin \\theta - \\cos \\theta ) \\left( \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta + \\sin \\theta \\cos \\theta \\right) } { \\sin \\theta \\cos \\theta ( \\sin \\theta - \\cos \\theta ) }{/tex}{tex}\\left[ {\\because {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}} \\right){/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta } + \\frac { \\cos ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta } + \\frac { \\sin \\theta \\cos \\theta } { \\sin \\theta \\cos \\theta }{/tex}{tex}= \\tan \\theta + \\cot \\theta + 1 = 1 + \\tan \\theta + \\cot \\theta = R . H S \\text { proved }{/tex}Since,\xa0{tex}\\tan A = \\frac{{\\sin A}}{{\\cos A}}{/tex}{tex}\\cot A = \\frac{{\\cos A}}{{\\sin A}}{/tex} | |
| 21934. |
if a and b are the zeros of the quadratic polynomial ax2+ bx+ c then evalute a-b |
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Answer» How u post image ax^2 and a and b are α,β |
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| 21935. |
a³+b³=? |
| Answer» (a+b) (a square + b square + a*b) | |
| 21936. |
The ratio of 7th to the 3rd term of an AP is 12:5. Find the ratio of 13th to the 4th term. |
| Answer» | |
| 21937. |
Odinate point on x axis |
| Answer» Any no. X and y=0 therefore (x,0) | |
| 21938. |
If (x-2)(x+3) hcf of polynomialp(x)=(x2-3x+2)(2x2+7x+a)q(x)=(x2+4x+3)(3x2-7x+b)Find a and b |
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| 21939. |
What is globilasation |
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Answer» Globalisation is the rapid integration and interconnection between countries Interconnection and integration between countries Interconnection and integration of countries by means of commercial purposes. Globalisation is a process of rapid integration or interconnection between countries |
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| 21940. |
The sum of integer is reciprocal 145/12 .find integares |
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Answer» How ram yuo posted this pic |
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| 21941. |
Solve for x;1/x+4-1/x-7=11/30 ,x not equal to-4,7 |
| Answer» | |
| 21942. |
Pgt converse |
| Answer» Pythagoras converse......?....... refer your NCERT it is too long to explain here nd most important.......We can\'t draw a figure here? | |
| 21943. |
In triangle abc angle c=3 angle b=2(angle a+anlge b). Find the 3 angles. |
| Answer» 20,40,120 in degrees. | |
| 21944. |
Rute 2 |
| Answer» 1.414 | |
| 21945. |
(x-1)-5(x-1)-6 |
| Answer» X= 1/2 | |
| 21946. |
Yadi dhup m khade wyakti ki chhcya uski uchai ki √3guna ho to, us samay surya k unnayan kon kya hoga |
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Answer» Not understand what you ask 30° Sorry but my maths language is english......but yours is hindi so it\'s not easy for me to solve your question well again sorry but I hope someone will give u answer soon... ???.... Not able to understand ur question |
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| 21947. |
If tanA+sinA=m & tanA -sinA=n show that m²-n²=4√mn |
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Answer» Solve rhs and lhs separately by putting the value of m and n Ya, I got the answer Bro.. |
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| 21948. |
4/x+3y=14 and 3/x-4y=23 |
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Answer» Sorry x=4/20=1/5 and y=-2 {tex}\\begin{array}{l}\\frac4{\\mathrm x}+3\\mathrm y=14\\;\\;-----(1)\\\\\\frac3{\\mathrm x}-4\\mathrm y=23\\;-----(2)\\\\\\mathrm{Multiply}\\;(1)\\;\\mathrm{by}\\;3\\;\\mathrm{and}\\;(2)\\;\\mathrm{by}\\;(4)\\;\\mathrm{we}\\;\\mathrm{get}\\\\\\frac{12}{\\mathrm x}+9\\mathrm y=42----(3)\\\\\\frac{12}{\\mathrm x}-16\\mathrm y=92\\;----(4)\\\\\\mathrm{Subtracting}\\;(4\\;\\mathrm{from}(3)\\mathrm{we}\\;\\mathrm{get}\\\\0\\;+25\\mathrm y=42-92=-50\\\\\\mathrm{So}\\;\\;\\mathrm y=-50/25=-2\\\\\\\\\\\\\\end{array}{/tex}{tex}\\begin{array}{l}\\mathrm{puttig}\\;\\mathrm{value}\\;\\mathrm{of}\\;\\mathrm y\\;\\mathrm{in}\\;(\\;1)\\;\\mathrm{we}\\;\\mathrm{get}\\\\\\;4/\\mathrm x+3\\mathrm y=14\\;\\\\\\;4/\\mathrm x+3(-2)=14\\\\4/\\mathrm x-6=14\\;\\\\4/\\mathrm x=20\\;\\;\\;\\mathrm{So}\\;\\mathrm x=5/20=1/4\\;\\;\\;\\mathrm{and}\\;\\mathrm y=\\;-2\\;\\mathrm{ans}\\\\\\\\\\\\\\end{array}{/tex} x=0 and y=19/25 |
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| 21949. |
How to calculate theta in area of sector formula and in length of arc and in area of segment formula |
| Answer» If length of arc and radius is given then; Theta /360×2πr=length of arc | |
| 21950. |
2(ax-by) +(a+4b)=0 2(bx+ay)+(b-4a) =0 |
| Answer» Answer the question | |