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InterviewSolution
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.
| 42901. |
Tan theta/(sec theta-1)+Tan theta /(sec theta +1) =2cos theta |
| Answer» LHS ={(Sec theta +1)*tan theta + ( sec theta -1)*tan theta}/ (sec sq. theta -1) ={Tan theta (sec theta+1+sec theta-1)}/(tan theta*tan theta)=(2sec theta)/ tan theta =2Cosec theta =RHS=2 cosec theta =LHS=RHS: Hence proved | |
| 42902. |
Optional exercises need to be practiced or not |
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Answer» Noooooooo Yes but not for all chapters Thank you No |
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| 42903. |
Product of rational and irrational number is ?? |
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Answer» Irrational number Always irrational The product of rational and irrational number is an irrational number. It will be irrational .For example 2×√3=2√3 here 2 is rational ,√3 is irrational and 2√3 is which is the product of rational and irrational is also an irrational number. irrational |
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| 42904. |
Origin divides line joining of points (1;0) and (-1;-1 ) in ratio of which ? |
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Answer» Hiiii 2:3 |
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| 42905. |
The Perimeter of triangle formed by the points (0,0),(1,0)and (0,1). |
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Answer» It forms a right triangle with base 1 unit and altitude 1 unit so hypotaneous becomes (√2)Hence perimeter is (2 + √2)units. 1/2 |X1(y2-y3)+X2(y3- y1)+X3(y1-y2)|1/2|0(0-1)+1(1-0)+0(0-0)|1/2|0+1+0|1/2 * 11/2 ans |
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| 42906. |
figure, DE||BC and AD =1 cm, BD = 2 m. The ratio of the area of to the area of |
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| 42907. |
Find x if distance between the points (x,2) and (3,4) be 8 units. |
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| 42908. |
(-6,-7),(-1,-5) |
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| 42909. |
Express 98 as a product of it\'s primes |
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Answer» 2×7×7 2*7*7 2×7×7 |
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| 42910. |
what is the difference between 10th class basic and standard paper? |
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Answer» In 10th class basic paper questions comes from NCERT Book70% and for standard paper questions become ncert base which is only 30% Standard mathematics is comparitively hard as compared to basic math, if you choose basic math you can\'t take science stream and few courses in commerce field,whereas if we take standard mathematics all choices and streams are open for you @Likhit R VERY CORRECT!! Standard mathematics is comparitively hard as compared to basic math, if you choose basic math you can\'t take science stream and few courses in commerce field,whereas if we take standard mathematics all choices and streams are open for you I also DK? |
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| 42911. |
4s2 - 4s + 1 |
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Answer» 4s2-2s-2s+1=2s(2s-1)-1(2s-1)=(2s-1)(2s-1)=1\\2,1\\2 4s2- 2s -2s +1= 2s(2s-1)-1(2s-1)=(2s-1)(2s-1) S= 1/2 Kskdjrodur dfkdlw |
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| 42912. |
Square root of 49 |
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Answer» √49 = √7×7 = 7 7 7*7 = 49So the √49 is 7 7 √49 = 7 |
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| 42913. |
Explain triangles please |
| Answer» Triangle means a shape made up of three side or three straight line In triangle their are two types:-1.sides. 2.angleSIDES:-. ANGLE:-¹.equilateral ¹.acute².isosceles ².obtuse³.scalen. ³.rightSimilarities property:-RHSAASSASSSSA triangle are said to be similar when their sides are proportional and opposite angle are equal . | |
| 42914. |
The shape of the graph of a cubic polynomial is |
| Answer» Triangle | |
| 42915. |
How many revolution does a wheel made to cover a distance of 352m when radius of wheel is 0.7m |
| Answer» 5©®0®€$ | |
| 42916. |
Experience 7.1 |
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Answer» Hi Questions no 1 |
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| 42917. |
How 45=90 ??? |
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Answer» 45× 2 = 90 45×2 = 90 (45)^2=90 45*2 = 90 |
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| 42918. |
Use Euclid\'s division algorithm to find the HCF of 135 and 225 |
| Answer» By using Euclid\'s division lemma(a=bq+r)WhereA=225. B=135Step 1225=135×1+90Step 2A=135. B=90135=90×1+45Step 3A=90. B=4590=45×2+0So our HCF of (135 and 225) is 45 Hence proved | |
| 42919. |
If x cosA=8 and 15 cosecA =8secA then the value of x is? |
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Answer» ??Wow bro ?????? Edited table : $$\\sf \\color{aqua}{Trigonometry\\: Table}\\\\ \\blue{\\boxed{\\boxed{\\begin{array}{ |c |c|c|c|c|c|} \\sf \\red{\\angle A} & \\red{\\sf{0}^{ \\circ} }&\\red{ \\sf{30}^{ \\circ} }& \\red{\\sf{45}^{ \\circ} }& \\red{\\sf{60}^{ \\circ}} &\\red{ \\sf{90}^{ \\circ}} \\\\ \\hline \\\\ \\rm \\red{sin A} & \\green{0} & \\green{\\dfrac{1}{2}}& \\green{\\dfrac{1}{ \\sqrt{2} }} &\\green{ \\dfrac{ \\sqrt{3}}{2} }&\\green{1} \\\\ \\hline \\\\ \\rm \\red{cos \\: A} & \\green{1} &\\green{ \\dfrac{ \\sqrt{3} }{2}}&\\green{ \\dfrac{1}{ \\sqrt{2} }} & \\green{\\dfrac{1}{2}} &\\green{0} \\\\ \\hline \\\\\\rm \\red{tan A}& \\green{0} &\\green{ \\dfrac{1}{ \\sqrt{3} }}&\\green{1} & \\green{\\sqrt{3}} & \\rm \\green{\\infty} \\\\ \\hline \\\\ \\rm \\red{cosec A }& \\rm \\green{\\infty} & \\green{2}& \\green{\\sqrt{2} }&\\green{ \\dfrac{2}{ \\sqrt{3} }}&\\green{1} \\\\ \\hline\\\\ \\rm \\red{sec A} & \\green{1 }&\\green{ \\dfrac{2}{ \\sqrt{3} }}& \\green{\\sqrt{2}} & \\green{2} & \\rm \\green{\\infty} \\\\ \\hline \\\\ \\rm \\red{cot A }& \\rm \\green{\\infty} & \\green{\\sqrt{3}}& \\green{1} & \\green{\\dfrac{1}{ \\sqrt{3} }} & \\green{0}\\end{array}}}}$$ $$ \\sf \\bf :longmapsto xcosA = 8$$$$\\sf \\bf :longmapsto cosA=\\dfrac{8}{x}$$Taking second equation,$$\\sf \\bf :longmapsto 15cosecA = 8secA$$$$\\sf \\bf :longmapsto 15\\dfrac{1}{sinA} = 8\\dfrac{1}{cosA}$$$$\\sf \\bf :longmapsto tanA = \\dfrac{15}{8}$$Thus, you can see using a triangle that sides are respectively, 15k, 8k, 17k$$\\sf \\bf :longmapsto cosA = \\dfrac{8}{17}$$Hence , x = 17More information:[tex]\\sf \\color{aqua}{Trigonometry\\: Table}\\\\ \\blue{\\boxed{\\boxed{\\begin{array}{ |c |c|c|c|c|c|} \\sf \\red{\\angle A} & \\red{\\sf{0}^{ \\circ} }&\\red{ \\sf{30}^{ \\circ} }& \\red{\\sf{45}^{ \\circ} }& \\red{\\sf{60}^{ \\circ}} &\\red{ \\sf{90}^{ \\circ}} \\\\ \\hline \\\\ \\rm \\red{sin A} & \\green{0} & \\green{\\dfrac{1}{2}}& \\green{\\dfrac{1}{ \\sqrt{2} }} &\\green{ \\dfrac{ \\sqrt{3}}{2} }&\\green{1} \\\\ \\hline \\\\ \\rm \\red{cos \\: A} & \\green{1} &\\green{ \\dfrac{ \\sqrt{3} }{2}}&\\green{ \\dfrac{1}{ \\sqrt{2} }} & \\green{\\dfrac{1}{2}} &\\green{0} \\\\ \\hline \\\\\\rm \\red{tan A}& \\green{0} &\\green{ \\dfrac{1}{ \\sqrt{3} }}&\\green{1} & \\green{\\sqrt{3}} & \\rm \\green{\\infty} \\\\ \\hline \\\\ \\rm \\red{cosec A }& \\rm \\green{\\infty} & \\green{2}& \\green{\\sqrt{2} }&\\green{ \\dfrac{2}{ \\sqrt{3} }}&\\green{1} \\\\ \\hline\\\\ \\rm \\red{sec A} & \\green{1 }&\\green{ \\dfrac{2}{ \\sqrt{3} }}& \\green{\\sqrt{2}} & \\green{2} & \\rm \\green{\\infty} \\\\ \\hline \\\\ \\rm \\red{cot A }& \\rm \\green{\\infty} & \\green{\\sqrt{3}}& \\green{1} & \\green{\\dfrac{1}{ \\sqrt{3} }} & \\green{0}\\end{array}}}}[/tex] 17 |
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| 42920. |
Prove that the point (7,10), (-2,5)and (3,-4)are the vertices of an isosceles right triangle |
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Answer» You should find the distance between three pairs of vertices AB,BC,CA If two distances are equal and the points are not collinear then points form a isosceles triangle.To prove that it is also right triangle , use the converse of Pythagoras Check that is square of longest sides equal to sum of square of two shorter side.H^2=P^2+B^2 |
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| 42921. |
Triangle and trigonometry mera doubt hai |
| Answer» Hi | |
| 42922. |
2root3x^2-6x+root 3 |
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| 42923. |
SecA-1+Sec+1 |
| Answer» SecA -1+SecA+1=SecA+SecA=2SecA | |
| 42924. |
If 13 |
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| 42925. |
Trigonometry 8.4 5(5) |
| Answer» (v) (cos A – sin A + 1) / (cos A + sin A – 1) = cosec A + cot A using the identity cosec2A = 1 + cot2ASolving LHSMultiplying numerator and denominator by (cot A – 1 + cosec A)= (cot2A + 1 + cosec2A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot2A – (1 + cosec2A – 2*cosec A))= (2*cosec2A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot2A – 1 – cosec2A + 2*cosec A)\xa0= (2* cosec A *(cosec A + cot A) – 2*(cosec A + cot A)) / (cot2A – 1 – cosec2A + 2*cosec A)= ((cosec A + cot A) * (2*cosec A – 2 )) / (2*cosec A – 2)\xa0= cosec A + cot A = RHSHence Proved | |
| 42926. |
Displacement and rotation of a geometrical figure |
| Answer» 1.A cut out of a geometrical figure such as a triangle is made and placed on a rectangular sheet of paper marked with X and Y-axis.2.The co-ordinates of the vertices of the triangle and its centroid are noted.3.The triangular cut out is displaced(along x-axis, along y-axis or along any other direction.)4.The new co-ordinates of the vertices and the centroid are noted again.5.The procedure is repeated, this time by rotating the triangle as well as displacing it. The new co-ordinate of vertices and centroid are noted again. Displacement and rotation of a geometrical figure.6.Using the distance formula, distance between the vertices of the triangle are obtained for the triangle in original position and in various displaced and noted positions.7.Using the new coordinates of the vertices and the centroids, students will obtain the ratio in which the centroid divides the medians for various displaced and rotated positions of the triangles. | |
| 42927. |
The product of two consecutive positive integer is divided by 2 |
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Answer» 2 Thankyou Hence any integer is of one of the form 2q, 2q+1. Hence n(n+1) = 2((2q+1)(q+1)), which is even. Hence n(n+1) is always even. Hence the product of two consecutive integers is\xa0always divisible by\xa02. |
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| 42928. |
What is the distance covered by the tip of hour hand in 3.5 hours |
| Answer» Complete your question. | |
| 42929. |
N square - 1 is divisible by 8 if n is |
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Answer» An odd integer Odd integer Odd positive integer 3 also 5 |
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| 42930. |
P is a LCM of 2,4,6,8,10 and q is the LCM of 1,3,5,7,9 and l is the LCM of p&q |
| Answer» L=3(8) (5)21 or 5(7) (9) 8 | |
| 42931. |
The value of (1+tan |
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Answer» Please complete your question Please complete your q |
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| 42932. |
If a and b are the zeroes of polynomial f(x)=x2-3x+2 then 1/a and 1/b is equal to |
| Answer» p(x) = x² - 3x + 2=> x² - 2x - x + 2 = 0=> x(x - 2)-1(x - 2) = 0=> (x - 1)(x - 2) = 0=> x - 1 = 0, x = 1 => a=> x - 2 = 0, x = 2 => btherefore two zeros a and b are 1 and 21/a + 1/b = 1/1 + 1/2=> (2+1)/2=> 3/2. Answer | |
| 42933. |
Chennai sahodaya common pre board paper |
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| 42934. |
Chapter 12, exercise=12.3, Question7 |
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Answer» Area of square - Area of circle 42cm² Area of square-area of 4sector |
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| 42935. |
Sec^4A - Sec^2 A is equal to ? |
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Answer» Since all the options involve the trigonometric ratio tan θ, so we use the identity 1 + tan2θ = sec2θ. To find: sec4A – sec2A Consider sec4A – sec2A = (sec2A)2 – sec2A Now, as sec2A = 1 + tan2A ⇒ sec4A – sec2A = (sec2A)2 – sec2A = (1 + tan2A)2 – (1 + tan2A) = 1 + tan4A + 2 tan2A – 1 – tan2A = tan4A + tan2A Sec^4A-sec^2A =\xa0tan^4A+tan^2 |
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| 42936. |
HCF of 306 and 1314 |
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Answer» 18 18 18 is your answer |
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| 42937. |
cos 45°/ sec 30°+cosec 30°. Evaluate? |
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| 42938. |
Circumference of a circle exceeds its diameter by 180 cm then its radius is |
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| 42939. |
If a= xy^2 and b= x^3y^5 where x and y is prime number then its LCM is |
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Answer» X^3y^5 X^3y^5 Zxv x^3y^5 step by step |
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| 42940. |
What is the probability that a leap year selected at random will have 52 sundays |
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Answer» 1/7 1/7 is the write ans 2/7 1/52 |
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| 42941. |
3x^2+5x-2 plitting the term we get? |
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Answer» 3x^2+5x-23x^2+6x-x-23x(x+2)-(x+2)(x+2)(3x-1) 3x^2+6x-x-2 3x(x+2)-1(x+2) (3x-1) (x+2) 3x^2+6x-x-23x(x+2)-1(x+2)(3x-1)(x+2) |
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| 42942. |
12.3 q1 |
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| 42943. |
If the zeros of the polynomial F(x)=2x³-15x²+37x-30 are in A.P. ,find them |
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Answer» Ohk thanks ? By a-d,a,a+d 2,5/2,3 |
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| 42944. |
X2 + X2 = |
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Answer» X2 2X2 Accordingly the question that you have raised the coefficient of the polynomial are 1 hence the answer should be 2x raise to power2........2x^2 |
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| 42945. |
In the backyard of house parallel case study |
| Answer» Idk ?? | |
| 42946. |
Anyone know class10 term-1 board syllabus of all subject pls....tell me??? |
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Answer» Pls tell me Ss i now |
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| 42947. |
The length of diagonals of a rhombus are 24 cm and 36 cm, then find its attitude. |
| Answer» 19.2 | |
| 42948. |
The rational form of 0.25454545454545....... is in the form of p/q ,then (p+q)is |
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Answer» 0.2545454.... = 0.2(_54_) = xSo , 10x = 2._54_ and 1000x = 254._54_ .Then , 990x = 252 , x = 252/990x = 14/55 ; p+q = 69 Ans. 254/90 P/q |
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| 42949. |
A company manufacturers two type of bottles alpha and beta |
| Answer» Hi | |
| 42950. |
If tan theta root 3 then |
| Answer» Theta is 60 | |