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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.
| 17101. |
(X+1) (X+2) (X+3) (X+4) = 120 |
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Answer» Rewrite the equation as:(x + 1)(x + 4)(x + 3)(x + 2) = 120Multiply the first 2 and last 2 expressions:\xa0..... (1)Let\xa0(1) becomes,(y + 4)(y + 6) = 120y2\xa0+ 10x + 24 = 120y2\xa0+ 10x - 96 = 0y2\xa0+ 16x - 6y - 96 = 0y(y + 16) - 6(y + 16) = 0(y + 16)(y - 6) = 0y = -16 or 6y2+ 5x = 16 or 6When x2\xa0+ 5x = 6x2\xa0+ 5x - 6 = 0x2\xa0+ 6x - x - 6 = 0x(x + 6) - 1(x + 6) = 0(x + 6)(x - 1) = 0x = -6, 1When x^2 + 5x = -16x^2 + 5x + 16 = 0Using quadratic These are all the possible solutions. But the only real solutions are -6 and 1. Rewrite the equation as:(x + 1)(x + 4)(x + 3)(x + 2) = 120Multiply the first 2 and last 2 expressions:{tex}(x^2 + 5x + 4)(x^2 + 5x + 6) = 120{/tex} ..... (1)Let {tex}x^2 + 5x = y{/tex}(1) becomes,(y + 4)(y + 6) = 120y2 + 10x + 24 = 120y2 + 10x - 96 = 0y2 + 16x - 6y - 96 = 0y(y + 16) - 6(y + 16) = 0(y + 16)(y - 6) = 0y = -16 or 6y2+ 5x = 16 or 6When x2 + 5x = 6x2 + 5x - 6 = 0x2 + 6x - x - 6 = 0x(x + 6) - 1(x + 6) = 0(x + 6)(x - 1) = 0x = -6, 1When x^2 + 5x = -16x^2 + 5x + 16 = 0Using quadratic formula:{tex}x = [-5 +/- sqrt (25 - 64)]/2{/tex}The solutions are, -6, 1, [-5 - sqrt (25 - 64)]/2 and [-5 + sqrt (25 - 64)]/2These are all the possible solutions. But the only real solutions are -6 and 1. |
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| 17102. |
ABC is an equilateral triangle of side 2a. What is the ratio of its side to its altitude |
| Answer» 2:√3 | |
| 17103. |
Find the ratio in which the line segment joining (1,-7) and (6, 4) is divided by x-axis |
| Answer» Given :\xa0Line segment joining the points (6,4)and(1,-7) is divided by x axis.To find :\xa0The ratio in the line segment and the coordinates of point of division?Solution :Let the line segment points A=(6,4) and B=(1,-7)Let the line segment divide by x-axis with point P=(x,0)Let the ratio in which line segment divide be m:n=k : 1Applying section formula,Substitute the values,Compare the y-coordinate,So, The ratio is 4 :7.Compare the x-coordinate,Put the value of k,So, The coordinate of point of division is\xa0 | |
| 17104. |
Prove thata the tangent drawn at the end of the chord of a circle make equal angles with the chord |
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Answer» Yes , right answer Given:- A circle with center\xa0O,PA\xa0and\xa0PB\xa0are tangents drawn at ends\xa0A\xa0and\xa0B\xa0on chord\xa0AB.To prove:-\xa0∠PAB=∠PBAConstruction:- Join\xa0OA\xa0and\xa0OBProof:- In\xa0△AOB, we haveOA=OB(Radii\xa0of\xa0the\xa0same\xa0circle)∠OAB=∠OBA.....(1)(Angles\xa0opposite\xa0to\xa0equal\xa0sides)∠OAP=∠OBP=90(∵Radius⊥Tangent)⇒∠OAB+∠PAB=∠OBA+∠PBA⇒∠OAB+∠PAB=∠OAB+∠PBA(From\xa0(1))⇒∠PAB=∠PBAHence proved. |
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| 17105. |
2sinA = sin2A when A? |
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Answer» A=0° sin2A=2sinA2sinA.cosA-2sinA =02sinA(cosA-1)=0either 2sinA=0sinA=0sinA=sin0°A= 0°or cosA -1=0cosA=1cosA=cos 0°A=0° , Answer |
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| 17106. |
Odd number always end in one of the digits ______ |
| Answer» Odd number always end in one of the digits\xa01, 3, 5, 7, 9 | |
| 17107. |
Theta /360 *π®2 =1155/8 radius 21/2 find theta |
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Answer» 150 degree 60° |
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| 17108. |
1_x + 1_y= 5 |
| Answer» Please visit my channel cheersome Manya on YouTube .Please support me by subscribing | |
| 17109. |
What is the full form of A.P |
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Answer» Algorithiem propagation arithmetic progression Arithmetic progression The full form of A.P is Arithmetic progression Which subject ??? |
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| 17110. |
8x+5y=93x+2y=4Solve using elimination method |
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Answer» ANSWERMultiply the equation\xa08x+5y=9\xa0by\xa03\xa0and\xa0equation\xa03x+2y=4\xa0by\xa03\xa0to make the coefficients of\xa0x\xa0equal. Then we get the equations:24x+15y=27.........(1)24x+16y=32.........(2)Subtract Equation (1) from Equation (2) to eliminate\xa0x, because the coefficients of\xa0x\xa0are the same. So, we get(24x−24x)+(16y−15y)=32−27i.e.\xa0y=5Substituting this value of\xa0y\xa0in\xa0the equation\xa08x+5y=9,\xa0we get8x+25=9i.e.\xa08x=−16i.e.\xa0x=−2Hence, the solution of the equations is\xa0x=−2,y=5. Let 1 and 2 be equation 1 and 2 respectively Multiplying equation 1 by 216x+ 10y = 18---- equation 3Multiplying equation 2 by 515x +10y =20------ equation 4equation 3-equation 416x+10y=18-(+15x+10y=20)___________________x=-2Putting value of x in equation 316(-2)+10y=18-32+10y=1810y=18+3210y=50Y=50÷10Y=10 |
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| 17111. |
What is the portion of maths??? |
| Answer» 1) polynomials2) pair of linear equation in two varible3) quadratic equation4) Arithmetic progressions5) coordinate geometry6) surface areas and volume7) statics8) probablityI don\'t know about other chapter\'s this much only I know. | |
| 17112. |
Exercise 4.2 question number 2 |
| Answer» 2. Write four solutions for each of the following equations:(i) 2x+y = 7Solution:To find the four solutions of 2x+y =7 we substitute different values for x and yLet x = 0Then,2x+y = 7(20)+y = 7y = 7(0,7)Let x = 1Then,2x+y = 7(2×1)+y = 72+y = 7y = 7-2y = 5(1,5)Let y = 1Then,2x+y = 7(2x)+1 = 72x = 7-12x = 6x = 6/2x = 3(3,1)Let x = 2Then,2x+y = 7(2×2)+y = 74+y = 7y =7-4y = 3(2,3)The solutions are (0, 7), (1,5), (3,1), (2,3)(ii) πx+y = 9Solution:To find the four solutions of πx+y = 9 we substitute different values for x and yLet x = 0Then,πx+y = 9(π0)+y = 9y = 9(0,9)Let x = 1Then,πx +y = 9(π×1)+y = 9π+y = 9y = 9-(1, 9-)Let y = 0Then,πx+y = 9πx+0 = 9πx = 9x = 9/(9/,0)Let x = -1Then,πx + y = 9(×-1) + y = 9-+y = 9y = 9+π(-1,9+)The solutions are (0,9), (1,9-), (9/,0), (-1,9+)(iii) x = 4ySolution:To find the four solutions of x = 4y we substitute different values for x and yLet x = 0Then,x = 4y0 = 4y4y= 0y = 0/4y = 0(0,0)Let x = 1Then,x = 4y1 = 4y4y = 1y = 1/4(1,1/4)Let y = 4Then,x = 4yx= 4×4x = 16(16,4)Let y =Then,x = 4yx = 4×1x = 4(4,1)The solutions are (0,0), (1,1/4), (16,4), (4,1) | |
| 17113. |
If the third and the ninth term of an AP are 4 and 8 respectively,which term of this AP is 0? |
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Answer» Yogita a great job... an\u200b=a+(n−1)da3\u200b=4=a+2d\xa0....(1)a9\u200b=−8=a+8d\xa0....(2)∴4−2d=−8d−8−2d+8d=−8−46d=−12d=−2\u200ba=4−2d=4+4=8an\u200b=0,a=8,d=−20=8+(n−1)×(−2)0=8−2n+22n=10n=5 |
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| 17114. |
How many three digit nunbers are divisible by 7 ? |
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Answer» The first 3-digit number which is divisible by 7 is 105The last 3-digit number which is divisible by 7 is 994The list of 3-digit numbers divisible by 7 are105, 112, 119,…..994 which forms an A.PConsider a formulaT(n) = a + (n – 1)dWherea = 105d = 7T(n) = 994994 = 105 + (n – 1)7889 = 7n – 77n = 896n = 128∴ There are 128 3-digits number which are divisible by 7. 128 3∴ There are\xa0128 3-digits number which are divisible by 7. |
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| 17115. |
tanA/(1-cotA) +cotA/(1-tanA) =1+tanA+cotA |
| Answer» | |
| 17116. |
(1/x+1)+(2/x+2)=(4/x+4) find the roots. |
| Answer» https://www.shaalaa.com/question-bank-solutions/solve-x-1-x-1-2-x-2-4-x-4-x-1-2-3-nature-roots_3618#z=57ZopElj | |
| 17117. |
Prove that sinQ+1-cosQ/cosQ-1+sinQ=1+sinQ/cosQ |
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Answer» Par Aapne isme=ke baad wala kyu nhi Likha h Solution:LHS =\xa0Divide numerator and denominator by\xa0cosQ\xa0,\xa0we get=\xa0=/* By Trigonometric identity:Sec²Q-tan²Q = 1 */=\xa0=\xa0=\xa0=\xa0After cancellation, we get=\xa0Multiply\xa0numerator\xa0and\xa0denominator\xa0by\xa0(secQ-tanQ),\xa0we\xa0get=\xa0=\xa0=\xa0= RHS |
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| 17118. |
Prove that a parallelogram is a rhombus |
| Answer» \t\tIf all sides of a quadrilateral are congruent, then it’s a rhombus (reverse of the definition).\t\t\tIf the diagonals of a quadrilateral bisect all the angles, then it’s a rhombus (converse of a property).\t\t\tIf the diagonals of a quadrilateral are perpendicular bisectors of each other, then it’s a rhombus (converse of a property).\tTip:\xa0To visualize this one, take two pens or pencils of different lengths and make them cross each other at right angles and at their midpoints. Their four ends must form a diamond shape — a rhombus.\t\t\tIf two consecutive sides of a parallelogram are congruent, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).\t\t\tIf either diagonal of a parallelogram bisects two angles, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).\t\t\tIf the diagonals of a parallelogram are perpendicular, then it’s a rhombus (neither the reverse of the definition nor the converse of a property).\tHere’s a rhombus proof for you. Try to come up with a game plan before reading the two-column proof.Statement 1:Reason for statement 1:\xa0Given.Statement 2:Reason for statement 2:\xa0Opposite sides of a rectangle are congruent.Statement 3:Reason for statement 3:\xa0Given.Statement 4:Reason for statement 4:\xa0Like Divisions Theorem.Statement 5:Reason for statement 5:\xa0All angles of a rectangle are right angles.Statement 6:Reason for statement 6:\xa0All right angles are congruent.Statement 7:Reason for statement 7:\xa0Given.Statement 8:Reason for statement 8:\xa0A midpoint divides a segment into two congruent segments.Statement 9:Reason for statement 9:\xa0SAS, or Side-Angle-Side (4, 6, 8)Statement 10:Reason for statement 10:\xa0CPCTC (Corresponding Parts of Congruent Triangles are Congruent).Statement 11:Reason for statement 11:\xa0Given.Statement 12:Reason for statement 12:\xa0If a triangle is isosceles, then its two legs are congruent.Statement 13:Reason for statement 13:\xa0Transitivity (10 and 12).Statement 14:Reason for statement 14:\xa0If a quadrilateral has four congruent sides, then it’s a rhombus. | |
| 17119. |
Exersice 12.1 ka 3 no |
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Answer» Thanks ? The radius of 1st\xa0circle, r1\xa0= 21/2 cm (as diameter D is given as 21 cm)So, area of gold region = π r12\xa0= π(10.5)2\xa0= 346.5 cm2Now, it is given that each of the other bands is 10.5 cm wide,So, the radius of 2nd\xa0circle, r2\xa0= 10.5cm+10.5cm = 21 cmThus,∴ Area of red region = Area of 2nd\xa0circle − Area of gold region = (πr22−346.5) cm2= (π(21)2\xa0− 346.5) cm2= 1386 − 346.5= 1039.5 cm2Similarly,The radius of 3rd\xa0circle, r3\xa0= 21 cm+10.5 cm = 31.5 cmThe radius of 4th\xa0circle, r4\xa0= 31.5 cm+10.5 cm = 42 cmThe Radius of 5th\xa0circle, r5\xa0= 42 cm+10.5 cm = 52.5 cmFor the area of nth\xa0region,A = Area of circle n – Area of circle (n-1)∴ Area of blue region (n=3) = Area of third circle – Area of second circle= π(31.5)2\xa0– 1386 cm2= 3118.5 – 1386 cm2= 1732.5 cm2∴ Area of black region (n=4) = Area of fourth circle – Area of third circle= π(42)2\xa0– 1386 cm2= 5544 – 3118.5 cm2= 2425.5 cm2∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle= π(52.5)2\xa0– 5544 cm2= 8662.5 – 5544 cm2= 3118.5 cm2 |
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| 17120. |
If two vertices of a equilateral triangle be (0,0).(3,√3) find the third vertices |
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Answer» Two vertices of an equilateral triangle are (0, 0) and (3, √3).Let the third vertex of the equilaterla triangle be (x, y)Distance between (0, 0) and (x, y) = Distance between (0, 0) and (3, √3) = Distance between (x, y) and (3, √3)√(x2\xa0+ y2) = √(32\xa0+ 3) = √[(x - 3)2\xa0+ (y - √3)2]x2\xa0+ y2\xa0= 12x2\xa0+ 9 - 6x + y2\xa0+ 3 - 2√3y = 1224 - 6x - 2√3y = 12- 6x - 2√3y = - 123x + √3y = 6x = (6 - √3y) / 3⇒ [(6 - √3y)/3]2\xa0+ y2\xa0= 12⇒ (36 + 3y2\xa0- 12√3y) / 9 + y2\xa0= 12⇒ 36 + 3y2\xa0- 12√3y + 9y2\xa0= 108⇒ - 12√3y + 12y2\xa0- 72 = 0⇒ -√3y + y2\xa0- 6 = 0⇒ (y - 2√3)(y + √3) = 0⇒ y = 2√3 or - √3If y = 2√3, x = (6 - 6) / 3 = 0If y = -√3, x = (6 + 3) / 3 = 3So, the third vertex of the equilateral triangle = (0, 2√3) or (3, -√3). (0,4) |
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| 17121. |
1+1=5 |
| Answer» Hcf of 224 | |
| 17122. |
hindi mai maths |
| Answer» | |
| 17123. |
How many types of method to find the zeroes of quadratic equation |
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Answer» 3 formula Quadratic formula, factorisation , discriminate There is only one method for finding the zeros of quadratic equations and that is splitting the middle term process |
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| 17124. |
At 11o clock ,I send you a math class link |
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Answer» Yes maths class is going on Already maths class is going on . ??? |
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| 17125. |
ಈಗ ೨೨೩೫ನ್ನು ೪೩೫೬ರರಿಂದ ಗುಣಿಸಿ ಭಾಗಿಸಿದರೆ ಉತ್ತರ ಎಷ್ಟು ಬರುತ್ತದೆ |
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Answer» ಲೋ ಪಂಡಿತ್ತು ಯಾಕೋ ಕಂದ ಹಳೆ ಪ್ರಶ್ನೆಯಲ್ಲಿ ಏನೋ ಕೇಳಿದ್ದೇ ಯಾವ್ ಭಾಶೆ ಗೊತ್ತಾಯ್ತಾ ಏನು ಗೊತ್ತಾಗಿಲ್ಲ ಅಂದ್ರೆ ಹೇಳಲೇ ಸುಮೇಶ್ ಪಂಡಿತ್ ನಿನಗೆ ಕನ್ನಡ ಗೊತ್ತಿಲ್ವೇನಲೇ Divide 2235 by 4356 and then multiply their product with 500 ಯಾರಾದರೂ ಹೇಳಿ |
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| 17126. |
Express 156 as the product of primes. |
| Answer» 156 = 2 × 2 × 3 × 13 = 22\xa0× 3 × 13 | |
| 17127. |
use euclid ,s algorithm to find hcf (196,38220) |
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Answer» 38220>196 we always divide greater number with smaller one.Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as38220 = 196 * 195 + 0As there is no remainder so deviser 196 is our HCF 2+2 |
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| 17128. |
If sin A=¾, calculate cos A and tan A. |
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Answer» S o l u t i o nGiven\xa0sinA=3/4Or p/h=3/4Let p=3k and h=4kNow By Pythagoras theoremp2\xa0+ b2\xa0=h29k2\xa0+ b2\xa0=16k2b2\xa0= 7k2Or b=+k√7Now\xa0CosA=b/h=√7/4Now\xa0tanA=SinA/CosA=3/√7 |
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| 17129. |
Please send me a better time table so that i start my exam preparation from begining please |
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Answer» Instead of studying 3hrs study 1hr with full concentration it\'s more than enough than anythingDON\'T STUDY HARD,"STUDY SMART" Studying for 3hrs a day is enough if your full mind is on track I think so, according to me you should not make a timetable on timming like going to play at 6pm taking rest at 3pm "no" this is not a way you\'ll not follow it only for one or two days and then it will be hanged or pasted on wall unnecessary wasting your time in making charts just make a timetable like Monday Tuesday you will learn sst Wednesday Thursday English Friday Saturday Hindi and maths and science on regular bases as they are very important subjectsHope you will like it |
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| 17130. |
Find the value of (3^-1 + 4^-1)^0 + 5^-1 |
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Answer» 0.2 DIY and leave a thanks now it\'s not request 0.2 |
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| 17131. |
Write the coordinate of a point on x-axis which is equidistant from the points (-3,4) and (7,6). |
| Answer» Let the point on the x-axis be\xa0(x,0)Distance between\xa0(x,0)\xa0and\xa0(7,6)=√(7−x)2+(6−0)2\u200b=√72+x2−14x+36 \u200b= √x2−14x+85\u200bDistance between\xa0(x,0)\xa0and\xa0(−3,4)=√(−3−x)2+(4−0)2\u200b\xa0= √32+x2+6x+16\u200b= √x2+6x+25\u200bAs the point\xa0(x,0)\xa0is equidistant from the two points, both the distancescalculated are equal.x2−14x+85\u200b=x2+6x+25\u200b=>x2−14x+85=x2+6x+2585−25=6x+14x60=20xx=3Thus, the point is\xa0(3,0) | |
| 17132. |
Alpha+beta= |
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Answer» Yes alpha+beta=-b/a Also written as -b/a -(cofficient of x)/cofficient of x2 alpha +beta |
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| 17133. |
Solution of ex 3.3 question 3.(4) |
| Answer» The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?Solution:Let the fixed charge be Rs x and per km charge be Rs y.According to the question,x + 10y = 105 …………….. (1)x + 15y = 155 …………….. (2)From (1), we get x = 105 – 10y ………………. (3)Substituting the value of x in (2), we get105 – 10y + 15y = 1555y = 50y = 10 …………….. (4)Putting the value of y in (3), we getx = 105 – 10 × 10 = 5Hence, fixed charge is Rs 5 and per km charge = Rs 10Charge for 25 km = x + 25y = 5 + 250 = Rs 255 | |
| 17134. |
Tan theta = ? |
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Answer» What is the formula of cylinder Perpendicular /base P/B and sinA/cosA P/b P|b |
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| 17135. |
(Sin 30 + cos 30) - (sin 60 + Cos 60) |
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Answer» = [Sin30 + sin (90-30)] - [sin60 - sin (90 - 60 ) ]= sin 30 + sin 60 - sin 60 - sin 30=0 (Sin 30 + cos 30) - (sin 60 + Cos 60)= [Sin30 + sin (90-30)] - [sin60 - sin (90 - 60 ) ]= sin 30 + sin 60 - sin 60 - sin 30=0 |
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| 17136. |
Prove that 11×6+12×5+36 is a composite number |
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Answer» = 6 (27) = 162Therefore according to fundamental theorem of arithmetic product of a composite and prime number is always composite number. 11×6+12×5+36= 6(11× 1 + 2\xa0× 5 + 6)= 6(11 +10 + 6)= 6 (27) = 162Therefore according to fundamental theorem of arithmetic product of a composite and prime number is always composite number. |
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| 17137. |
If the sum of the zeroes of the quadratic polynomial 3x2-kx+6 is 3, then find the value of k |
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Answer» a =bq+r sum of roots=3 = -b/a ......1) 3x^2-kx+6 a=3,b= -k so, from 1equation -b/a= 3= -(-k)/3 3×3= +k 9=k answeranswer is 9 answer is 9 sum of roots=3 = -b/a ......1)3x^2-kx+6a=3,b= -kso, from 1equation-b/a= 3= -(-k)/33×3= +k9=k answer |
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| 17138. |
Is alternate segment theorem in class 10 syllabus |
| Answer» No | |
| 17139. |
4/7÷-7/4? |
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Answer» 4/7 *7/44/7x4/716/49 -16/49 Quadratic formula pz drive -1 is the answer. ............................ Solution: 4/7÷-7/4= -28/28= -1 |
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| 17140. |
Formula of area of quadrant of a circle? |
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Answer» 1/4 area of circle 1 by 4 pi r Square If any have doubts ask from me of all sibjects 1/4 πr^2 |
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| 17141. |
Chp - Introduction to trigonometry Find the value of 9 sec ² Φ - 9 tan²Φ |
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Answer» 9 9 is correct answer.. 9 1 |
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| 17142. |
Find the value of X, when in the A. P give below 2+6+10..........+X=1800 |
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Answer» Okk bro to advise Aapka math kamjor hai to isliye iss question ka answer khud hi dhundo aur thoda dimak lagao |
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| 17143. |
23$:÷ eue |
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Answer» https://teams.microsoft.com/l/meetup-join/19%3acf4294ecafc348bb92e109398b3dbc74%40thread.tacv2/1600398087278?context=%7b%22Tid%22%3a%22ce08424f-7e85-4ba6-8ea9-faf7e759094a%22%2c%22Oid%22%3a%220ccc15eb-bcfa-4c8b-967a-9640a86099ba%22%7d 2×6+2÷78÷;÷,$ |
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| 17144. |
Find all the zeroes of 3x⁴-11x³-x²+33x-24 |
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Answer» Plz... Check it. You can also get the answer of the question through google Check it again Question may be not completed |
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| 17145. |
15m+3n=66m+3n=1 |
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Answer» Right 15m+3n=6 ..... (i)6m+3n=1..... (ii)Subtract (i) and (ii) 15m + 3n = 6- 6m + 3n=1-------------------9m = 5m = 5/9Put m = 5/9 in (ii), we get6m + 3n = 16(5/9) + 3n = 110/3 + 3n = 13n = 1 - 10/33n = (3 - 10)/33n = -7/3n = -7/9 |
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| 17146. |
1-64/113=113-64 how |
| Answer» | |
| 17147. |
Secsq.x = 2+2tanx find tanx |
| Answer» | |
| 17148. |
Cos thita |
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Answer» Cos theta =B/H Cos theta = Sin(90°- theta) Cos theta = Sin(90°- theta) Cos theta = 1/Sec theta Cos thita = 1/sec thita |
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| 17149. |
Find the area of a sector of a circle with radius 6cm if angle of the sector is 60. |
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Answer» Working girl mu me te kar maruga 132/7cm^2 |
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| 17150. |
How can I proof Theorem 6.1 |
| Answer» Statement:\xa0If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.Given:\xa0A triangle ABC in which DE || BC, and intersects AB in D and AC in E. | |