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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.
| 401. |
Show that 1/2-√3 is irrational |
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| 402. |
Prove that 3√7(cube root 7) is an irrational |
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| 403. |
2x+2y=0 |
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Answer» x+y=o Or y=-x Linear equation in two variables Linear equation in two variables |
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| 404. |
What is the smallest LCM |
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Answer» 2 (two). HOPE IT HELPS YOU?✌ 2 "2". HOPE IT HELPS YOU?✌ 2 2 |
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| 405. |
Which term of AP: 121,117,113,.........,is its first negative term? |
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| 406. |
LCM OF 2 AND 4 |
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Answer» 4 The LCM of 2 and 4 is 4. 4 4 4 |
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| 407. |
Find the HCF of 128 and 268 |
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Answer» "4". HOPE IT HELPS YOU?✌ 4 4 4 4 |
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| 408. |
Find the value of sec 59 - cosec 31 |
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Answer» =sec(90-31)- cosec31 =cosec 31-cosec 31=0 0 0 |
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| 409. |
State and prove the Thales theorem (or B.P.T)? |
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Answer» See text book Base proportional theorem |
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| 410. |
Let a be any positive integer and b=4 solution |
| Answer» a+b=7 | |
| 411. |
All the formulas of maths of all chapter of class X cbse |
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Answer» Linear Equations One Variableax+b=0a≠0 and a&b are real numbersTwo variableax+by+c = 0a≠0 & b≠0 and a,b & c are real numbersThree Variableax+by+cz+d=0a≠0 , b≠0, c≠0 and a,b,c,d are real numbers Pair of Linear Equations in two variables: a1x+b1y+c1=0 a2x+b2y+c2=0 Where a1, b1, c1, a2, b2, and c2\xa0are all real numbers and a12+b12\xa0≠ 0 & a22\xa0+ b22\xa0≠ 0 It should be noted that\xa0linear equations in two variables\xa0can also be represented in graphical form. Algebra or Algebraic Equations The standard form of a Quadratic Equation is: ax2+bx+c=0 where a ≠ 0 And x = [-b ± √(b2\xa0– 4ac)]/2a Algebraic formulas: (a+b)2\xa0= a2\xa0+ b2\xa0+ 2ab (a-b)2\xa0= a2\xa0+ b2\xa0– 2ab (a+b) (a-b) = a2\xa0– b2 (x + a)(x + b) = x2\xa0+ (a + b)x + ab (x + a)(x – b) = x2\xa0+ (a – b)x – ab (x – a)(x + b) = x2\xa0+ (b – a)x – ab (x – a)(x – b) = x2\xa0– (a + b)x + ab (a + b)3\xa0= a3\xa0+ b3\xa0+ 3ab(a + b) (a – b)3\xa0= a3\xa0– b3\xa0– 3ab(a – b) (x + y + z)2\xa0= x2\xa0+ y2\xa0+ z2\xa0+ 2xy + 2yz + 2xz (x + y – z)2\xa0= x2\xa0+ y2\xa0+ z2\xa0+ 2xy – 2yz – 2xz (x – y + z)2\xa0= x2\xa0+ y2\xa0+ z2\xa0– 2xy – 2yz + 2xz (x – y – z)2\xa0= x2\xa0+ y2\xa0+ z2\xa0– 2xy + 2yz – 2xz x3\xa0+ y3\xa0+ z3\xa0– 3xyz = (x + y + z)(x2\xa0+ y2\xa0+ z2\xa0– xy – yz -xz) x2\xa0+ y2\xa0=½ [(x + y)2\xa0+ (x – y)2] (x + a) (x + b) (x + c) = x3\xa0+ (a + b +c)x2\xa0+ (ab + bc + ca)x + abc x3\xa0+ y3= (x + y) (x2\xa0– xy + y2) x3\xa0– y3\xa0= (x – y) (x2\xa0+ xy + y2) x2\xa0+ y2\xa0+ z2\xa0-xy – yz – zx = ½ [(x-y)2\xa0+ (y-z)2\xa0+ (z-x)2] Click here to check all algebra formulas Basic formulas for powers pm\xa0x pn\xa0= pm+n {pm}⁄{pn} = pm-n (pm)n\xa0= pmn p-m\xa0= 1/pm p1\xa0= p P0\xa0= 1 Arithmetic Progression(AP) Formulas If a1, a2, a3, a4, a5, a6,…\xa0are the terms of AP and d is the common difference between each term, then we can write the sequence as; a,\xa0a+d, a+2d, a+3d, a+4d, a+5d,….,nth term… where a is the first term. Now, nth\xa0term for\xa0arithmetic progression\xa0is given as; nth\xa0term = a + (n-1) d Sum of the first n terms\xa0in Arithmetic Progression; Sn\xa0= n/2 [2a + (n-1) d] Trigonometry Formulas For Class 10 Trigonometry maths formulas for Class 10 cover three major functions Sine, Cosine and Tangent for a right-angle triangle. Also, in\xa0trigonometry, the functions sec, cosec and cot formulas can be derived with the help of sin, cos and tan formulas. Let a right-angled triangle ABC is right-angled at point B and have\xa0∠θ. Sin θ=\xa0SideoppositetoangleθHypotenuse=PerpendicularHypotenuse\xa0= P/H Cos θ =\xa0AdjacentsidetoangleθHypotenuse\xa0=\xa0BaseHypotenuse\xa0= B/H Tan θ =\xa0SideoppositetoangleθAdjacentsidetoangleθ\xa0= P/B Sec θ =\xa01cosθ Cot θ =\xa01tanθ Cosec θ =\xa01sinθ Tan θ =\xa0SinθCosθ Trigonometry Table: Angle0°30°45°60°90°Sinθ01/21/√2√3/21Cosθ1√3/21/√2½0Tanθ01/√31√3UndefinedCotθUndefined√311/√30Secθ12/√3√22UndefinedCosecθUndefined2√22/√31 Swipe left Other Trigonometric formulas: sin(90°\xa0– θ) = cos θ cos(90°\xa0– θ) = sin θ tan(90°\xa0– θ) = cot θ cot(90°\xa0– θ) = tan θ sec(90°\xa0– θ) = cosecθ cosec(90°\xa0– θ) = secθ sin2θ + cos2\xa0θ = 1 sec2\xa0θ = 1 + tan2θ for 0°\xa0≤ θ < 90° Cosec2\xa0θ = 1 + cot2\xa0θ for 0°\xa0≤ θ ≤ 90° Get complete Trigonometry Formulas list here Circles Formulas For Class 10 Circumference of the circle = 2 π r Area of the circle = π r2 Area of the sector of angle θ = (θ/360) × π r2 Length of an arc of a sector of angle θ = (θ/360) × 2 π r (r = radius of the circle) Surface Area and Volumes Formulas For Class 10 The common formulas from the\xa0surface area and volumes\xa0chapter in 10th\xa0class include the following: Sphere Formulas Diameter of sphere2rSurface area of sphere4 π r2Volume of Sphere4/3 π r3 Cylinder Formulas Curved surface area of Cylinder2 πrhArea of two circular bases2 πr2Total surface area of CylinderCircumference of Cylinder + Curved surface area of Cylinder = 2 πrh + 2 πr2Volume of Cylinderπ r2\xa0h Cone Formulas Slant height of conel = √(r2\xa0+ h2)Curved surface area of coneπrlTotal surface area of coneπr (l + r)Volume of cone⅓ π r2\xa0h Cuboid Formulas Perimeter of cuboid4(l + b +h)Length of the longest diagonal of a cuboid√(l2\xa0+ b2\xa0+ h2)Total surface area of cuboid2(l×b + b×h + l×h)Volume of Cuboidl × b × h Here, l = length, b = breadth and h = height In case of Cube, put l = b = h = a, as cube all its sides of equal length, to find the surface area and volumes. Statistics Formulas for Class 10 In class 10, the chapter\xa0statistics\xa0mostly deals with finding the mean, median and mode of grouped data. (I) The mean of the grouped data\xa0can be found by 3 methods. Direct Method: x̅\xa0=\xa0∑ni=1fixi∑ni=1fi, where ∑fi\xa0xi\xa0is the sum of observations from value i = 1 to n And ∑fi\xa0is the number of observations from value i = 1 to n Assumed mean method\xa0:\xa0x̅\xa0=\xa0a+∑ni=1fidi∑ni=1fi Step deviation method : x̅\xa0=\xa0a+∑ni=1fiui∑ni=1fi×h (II) The mode of grouped data: Mode =\xa0l+f1–f02f1–f0–f2×h (III) The median for a grouped data: Median =\xa0l+n2–cff×h Linear EquationsOne Variableax+b=0a≠0 and a&b are real numbersTwo variableax+by+c = 0a≠0 & b≠0 and a,b & c are real numbersThree Variableax+by+cz+d=0a≠0 , b≠0, c≠0 and a,b,c,d are real numbersPair of Linear Equations in two variables:a1x+b1y+c1=0a2x+b2y+c2=0Wherea1, b1, c1, a2, b2, and c2\xa0are all real numbers anda12+b12\xa0≠ 0 & a22\xa0+ b22\xa0≠ 0It should be noted that\xa0linear equations in two variables\xa0can also be represented in graphical form.Algebra or Algebraic EquationsThe standard form of a Quadratic Equation is:ax2+bx+c=0 where a ≠ 0And x = [-b ± √(b2\xa0– 4ac)]/2aAlgebraic formulas:(a+b)2\xa0= a2\xa0+ b2\xa0+ 2ab(a-b)2\xa0= a2\xa0+ b2\xa0– 2ab(a+b) (a-b) = a2\xa0– b2(x + a)(x + b) = x2\xa0+ (a + b)x + ab(x + a)(x – b) = x2\xa0+ (a – b)x – ab(x – a)(x + b) = x2\xa0+ (b – a)x – ab(x – a)(x – b) = x2\xa0– (a + b)x + ab(a + b)3\xa0= a3\xa0+ b3\xa0+ 3ab(a + b)(a – b)3\xa0= a3\xa0– b3\xa0– 3ab(a – b)(x + y + z)2\xa0= x2\xa0+ y2\xa0+ z2\xa0+ 2xy + 2yz + 2xz(x + y – z)2\xa0= x2\xa0+ y2\xa0+ z2\xa0+ 2xy – 2yz – 2xz(x – y + z)2\xa0= x2\xa0+ y2\xa0+ z2\xa0– 2xy – 2yz + 2xz(x – y – z)2\xa0= x2\xa0+ y2\xa0+ z2\xa0– 2xy + 2yz – 2xzx3\xa0+ y3\xa0+ z3\xa0– 3xyz = (x + y + z)(x2\xa0+ y2\xa0+ z2\xa0– xy – yz -xz)x2\xa0+ y2\xa0=½ [(x + y)2\xa0+ (x – y)2](x + a) (x + b) (x + c) = x3\xa0+ (a + b +c)x2\xa0+ (ab + bc + ca)x + abcx3\xa0+ y3= (x + y) (x2\xa0– xy + y2)x3\xa0– y3\xa0= (x – y) (x2\xa0+ xy + y2)x2\xa0+ y2\xa0+ z2\xa0-xy – yz – zx = ½ [(x-y)2\xa0+ (y-z)2\xa0+ (z-x)2]Click here to check all algebra formulasBasic formulas for powerspm\xa0x pn\xa0= pm+n{pm}⁄{pn} = pm-n(pm)n\xa0= pmnp-m\xa0= 1/pmp1\xa0= pP0\xa0= 1Arithmetic Progression(AP) FormulasIf a1, a2, a3, a4, a5, a6,…\xa0are the terms of AP and d is the common difference between each term, then we can write the sequence as; a,\xa0a+d, a+2d, a+3d, a+4d, a+5d,….,nth term… where a is the first term. Now, nth\xa0term for\xa0arithmetic progression\xa0is given as;nth\xa0term = a + (n-1) dSum of the first n terms\xa0in Arithmetic Progression;Sn\xa0= n/2 [2a + (n-1) d]Trigonometry Formulas For Class 10Trigonometry maths formulas for Class 10 cover three major functions Sine, Cosine and Tangent for a right-angle triangle. Also, in\xa0trigonometry, the functions sec, cosec and cot formulas can be derived with the help of sin, cos and tan formulas.Let a right-angled triangle ABC is right-angled at point B and have\xa0∠θ.Sin θ=\xa0SideoppositetoangleθHypotenuse=PerpendicularHypotenuse\xa0= P/HCos θ =\xa0AdjacentsidetoangleθHypotenuse\xa0=\xa0BaseHypotenuse\xa0= B/HTan θ =\xa0SideoppositetoangleθAdjacentsidetoangleθ\xa0= P/BSec θ =\xa01cosθCot θ =\xa01tanθCosec θ =\xa01sinθTan θ =\xa0SinθCosθTrigonometry Table:Angle0°30°45°60°90°Sinθ01/21/√2√3/21Cosθ1√3/21/√2½0Tanθ01/√31√3UndefinedCotθUndefined√311/√30Secθ12/√3√22UndefinedCosecθUndefined2√22/√31Swipe leftOther Trigonometric formulas:sin(90°\xa0– θ) = cos θcos(90°\xa0– θ) = sin θtan(90°\xa0– θ) = cot θcot(90°\xa0– θ) = tan θsec(90°\xa0– θ) = cosecθcosec(90°\xa0– θ) = secθsin2θ + cos2\xa0θ = 1sec2\xa0θ = 1 + tan2θ for 0°\xa0≤ θ < 90°Cosec2\xa0θ = 1 + cot2\xa0θ for 0°\xa0≤ θ ≤ 90°Get complete Trigonometry Formulas list hereCircles Formulas For Class 10Circumference of the circle = 2 π rArea of the circle = π r2Area of the sector of angle θ = (θ/360) × π r2Length of an arc of a sector of angle θ = (θ/360) × 2 π r(r = radius of the circle)Surface Area and Volumes Formulas For Class 10The common formulas from the\xa0surface area and volumes\xa0chapter in 10th\xa0class include the following:Sphere FormulasDiameter of sphere2rSurface area of sphere4 π r2Volume of Sphere4/3 π r3Cylinder FormulasCurved surface area of Cylinder2 πrhArea of two circular bases2 πr2Total surface area of CylinderCircumference of Cylinder + Curved surface area of Cylinder = 2 πrh + 2 πr2Volume of Cylinderπ r2\xa0hCone FormulasSlant height of conel = √(r2\xa0+ h2)Curved surface area of coneπrlTotal surface area of coneπr (l + r)Volume of cone⅓ π r2\xa0hCuboid FormulasPerimeter of cuboid4(l + b +h)Length of the longest diagonal of a cuboid√(l2\xa0+ b2\xa0+ h2)Total surface area of cuboid2(l×b + b×h + l×h)Volume of Cuboidl × b × hHere, l = length, b = breadth and h = height In case of Cube, put l = b = h = a, as cube all its sides of equal length, to find the surface area and volumes.Statistics Formulas for Class 10In class 10, the chapter\xa0statistics\xa0mostly deals with finding the mean, median and mode of grouped data.(I) The mean of the grouped data\xa0can be found by 3 methods.Direct Method: x̅\xa0=\xa0∑ni=1fixi∑ni=1fi, where ∑fi\xa0xi\xa0is the sum of observations from value i = 1 to n And ∑fi\xa0is the number of observations from value i = 1 to nAssumed mean method\xa0:\xa0x̅\xa0=\xa0a+∑ni=1fidi∑ni=1fiStep deviation method : x̅\xa0=\xa0a+∑ni=1fiui∑ni=1fi×h(II) The mode of grouped data:Mode =\xa0l+f1–f02f1–f0–f2×h(III) The median for a grouped data:Median =\xa0l+n2–cff×h Chapter ya question ke photo bhejo thabhi to bta payenge Konsa ch ka ek ek kr k pucho aise to aap v confused hoge aur hm sb v |
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| 412. |
All the formulas of maths ofball chapter pf cbse board |
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| 413. |
Find the value of y for which the distance between point A(3,-1) and B(11,y) is 10 minutes |
| Answer» Join Betternation.xyz This is a site where u can have a lot of friends and fun ? ? | |
| 414. |
If that tan theta 15 /8 find the volume of the of all t ratio of theta book name r s agrwal |
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Answer» Solution: Tan theta =15/8(given)We know that,Tan theta=p/bNow, we use Pythagoras theoremSo,(Hypo)²=(perpendicular)²+(base)²Therefore,H²=(15)²+(8)² =225+64 = 289 H=√289 = 17Hence,Sin theta=15/17Cos theta=8/17Tan theta=15/8Cot theta=8/15Cosec theta=17/15Sec theta =17/8I hape it is helpful for you ch 10 ex 10 Question number 3 please help me |
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| 415. |
Show that square of any positive odd integer is of the form 8q+1, for some interger q |
| Answer» 8(8m2+14m+6)+1=8q+1 | |
| 416. |
Suggest me channel for the maths on YouTube |
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Answer» Mathematics class 10. Maths solver Maths teacher and maths padho sir Green board Green board and Mathematics class 10. |
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| 417. |
Find the hcf of 272 and 148 and express it as 272p+148q |
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Answer» How do we get the value of p and q \xa0272 = 2 × 2 × 2 × 2 × 17\xa0148 = 2 × 2 × 37Therefore ,HCF ( 272 , 148 ) = 2 × 2 = 4\xa0By using Euclid\'s algorithm :\xa0Since , 272 > 148 , we apply the division lemma to 272 and 148 , to get272 = 148 × 1 + 124\xa0Since the remainder 124 ≠ 0 , we apply the division lemma to 148 and 124 , to get148 = 124 × 1 + 24\xa0We consider the new divisor 124 and the new remainder 24 , and apply the division lemma to get124 = 24 × 5 + 4\xa0We consider the new divisor 24 and the new remainder 4 , and apply the division lemma to get24 = 4 × 6 + 0\xa0The remainder has now become zero , so our procedure stops , since the divisor at this stage is 4 , the HCF of 272 and 148 is 4 P=(6) , q=(-11) |
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| 418. |
If A(-5,7), B(-4,-5), C(-1,-6) and D(4,5) are the vertices of\xa0⊡ABCD. Find its area. |
| Answer» by joining B to D,you will get two Triangles ABD and BCD the area of triangle A b d1/2[-5(-5-5)+(-4)(5-7)+4(7+5)]=01[50+8+48]0×2=106/2=53square.unitsalso the area of triangle BCD=1/2[-4(-6-5)-1(5+5)+4(-5+6)]=1/2[44-10+4]=19 square.unitsarea of triangle ABC + area of triangle BCDso the area of quadrilateral ABCD =53+19=72square.units | |
| 419. |
Which term of AP is 21,41,63.....is 210. |
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Answer» In the given A.P ;a = 21d = 42-21 = 21l = 210n = ?l = a +(n-1)d210 = 21+(n-1)21n-1 = 189/21n = 9+1n = 10hence 210 is the 10th term of given A.P I think about it is not good question. Something is wrong in this question |
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| 420. |
How do we find mean by assume mean method in exercise 14.2question no.3 |
| Answer» Join Betternation.xyz This is a site where u can have a lot of friends and fun ? ? | |
| 421. |
Giouhoit |
| Answer» Join Betternation.xyz This is a site where u can have a lot of friends and fun ? ? | |
| 422. |
Factorise the Quadratic equations 2x²+x-300 |
| Answer» 2x^2 + 25x - 24x - 300x(2x+25)-12 (2x+25) (x-25) (2x+25) x=25 , x=-25/2 | |
| 423. |
Bnb |
| Answer» | |
| 424. |
Tan/seca+1 +tan/secA-1=2coseca |
| Answer» tanA/secA+1 + tanA/secA-1TanA(1/secA+1 + 1/secA-1) TanA (cosA/1+cosA + cosA/1-cosA) TanA {cosA(1/1+cosA + 1/1-cosA) TanA{cosA(1-cosA+1+cosA) /1-cos²A}TanA×2cosA/sin²ASinA/cosA×2cosA/sin²A2/sinA2cosecA | |
| 425. |
-5 is one the zeroes of 2x^2 +px -15, zeroes of p(x^2+x)+k are equal to each other. Find value of k |
| Answer» P(x)= 2x^2+px-15Zero is given -5P(-5)=0Than 2(-5)^2+p(-5)-15=0 50-5p-15=035=5p7=pPut value of p in p(x^2+x)+k 7x^2+7x+kBy using b^2-4ac=049-28k=0K=-49/-28=7/4Thanks | |
| 426. |
(1 + sinA - cosA) /(cosA + sinA - 1) ==( 1 + sinA) /(cosA) |
| Answer» | |
| 427. |
sin∧2 A cot∧2 A + cos∧2 A tan∧2 A = |
| Answer» Answer is 1.Because, sin∧2 A cot∧2 A + cos∧2 A tan∧2 A =(sin^2A*cos^2A/sin^2A) + (cos^2A*sin^2A/cos^2A).= cos^2A+sin^2ASo by indentity the answer is 1. | |
| 428. |
How we learn trigonometric ratio table? ... |
| Answer» | |
| 429. |
(420/x-8) - 420/x =3 write in the form of quadratic equation |
| Answer» (420/x-8) - 420/x LCM of denominator which is x^2 - 8x420x - 420x +3360=3x^2 -24x 3x^2 -24x -3360 quadratic equation I Hope it will help you. | |
| 430. |
2 If the graphs of the equations 2x + ay = 10 and 3x + y = 12 are parallel lines |
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| 431. |
Solve pair equation Q 1/2x + 1/3y = 2 |
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| 432. |
If tan A= cotB, prove that A+B =90° |
| Answer» Cot(90-A)=cotB90-A=BA+B=90 | |
| 433. |
a=5,d=3,an=50 find n and Sn. |
| Answer» Given,\xa0a=5,d=3,an\u200b=50⇒a+(n−1)d=50⇒5+(n−1)3=50⇒5+3n−3=50⇒3n=48⇒n=16∴S16\u200b=216\u200b[2a+(16−1)d]=8[2×5+15×3]=440Hence,\xa0n=16,S16\u200b=440 | |
| 434. |
Find the LCM and HCF of 180 and 144 by prime factorisation method.. |
| Answer» Factors of 144 = 2×2×2×2×3×3Factors of 180 = 2×2×3×3×5Factors of 192 = 2×2×2×2×2×2×3Common factors of 144, 180, 192 = {2,2,3}LCM (144,180,192) =2×2×2×2×2×2×3×3×5 = 2880So, HCF(144,180,192) = 2×2×3 = 12. | |
| 435. |
Is frequency polygon ( maths chapter 14 statistics ) there in board syllabus 2020??? |
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Answer» No Frequency polynomial is not there for the board syllabus Kabi khud be karla kro |
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| 436. |
Why did the visible man become invisible at the inn? What did he do then? |
| Answer» The invisible man (Griffin) first became visible after he slipped into a big London store for keeping warm and overslept there while wearing some clothes taken from the store. To escape, he removed his clothes, becoming invisible. Thus he became homeless and was wandering the streets of London. | |
| 437. |
Find the pair of linear equation and find there solution graphically |
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| 438. |
Find the sum of all integers between 100 and 500 which are divisible by 3 |
| Answer» | |
| 439. |
If alpha and beta are the zeros of the polynomial 3/2x^2+9x^2- 15 then find the 1/ alpha +1/beta |
| Answer» Highlighted | |
| 440. |
What is probability of having spade face cards |
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Answer» 1/13 4/52=1/13 |
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| 441. |
Find the area of the triangle whose base measure 24cm and the corresponding height measure 14.5 cm |
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| 442. |
(sec A + Tan A )(1-sin A)= |
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Answer» SecA = 1/cos a,tanA=sina /cosa(1/cos a+sina /cosa)(1_sina)1_sina)(1+sina)/cosa1_sin^2 a/cosaCos^2A/cosACosa Cos A |
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| 443. |
2009 board exam Questions |
| Answer» | |
| 444. |
The pairIf the zeros of quadratic polynomial ax |
| Answer» a.a.b.b | |
| 445. |
a railway half ticket cost half the full fare but the reservation charges are the same on half |
| Answer» | |
| 446. |
Deleted exercise and questions from maths according to new syllabus |
| Answer» Go on to google | |
| 447. |
Find out 2(9)3 |
| Answer» 1458 | |
| 448. |
Find the hcf of 96 and 404 by the prime factorisation method. Hence find there LCM. |
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Answer» The decimal experssion120/3*3*3*5*5*5*5*5*5*5is Since, 96 =\xa02\xa0× 2\xa0× 2\xa0× 2\xa0× 2\xa0× 3and, 404 =\xa02\xa0× 2\xa0× 101So, HCF of 96 and 404 =\xa0Product of common prime factors =\xa02\xa0× 2 =\xa04LCM =\xa02\xa0× 2\xa0× 2\xa0× 2\xa0× 2\xa0× 101\xa0× 3 =\xa09696 |
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| 449. |
Find the area of triangle whose sides are x=y, 3x =y , x+y =8 |
| Answer» 4 sq units. | |
| 450. |
What is the formula of LCM |
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Answer» This is the formula for LCM-Product of AB /HCF of AB = LCM Sorry LCM is 12 Least common multiple for example 6 , 12 LCM is 6 Product of AB /HCF of AB = LCM |
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