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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.
| 11601. |
Prove that altitude of triangle are concurrent |
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| 11602. |
The first 3 terms of an ap are respectively (3y-1), (3y+5) and (5y+1) find your. |
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Answer» In given AP,\xa0{tex}{a_2} - {a_1} = {a_3} - {a_2}{/tex}3y + 5 - 3y + 1 = 5y + 1 - 3y - 5=> 6 = 2y - 4=> y = 5Then the given AP will be 14, 20, 26, .....Here common difference is 20 - 14 = 6Thereforem the fourth term is 26 + 6 = 32 The first three terms (3y—1), (3y+5) and (5y+1) are in A.P, so the common difference will be sameThat is d1=d2(3y+5)—(3y—1) = (5y+1)—(3y+5)3y+5—3y+1=5y+1—3y—56=2y—46+4 =2y10=2y5=y Given an AP in whicha1 = 3y - 1a2 = 3y + 5a3 = 5y + 1we know that term 2 - term 1 = term 3 - term 2 = common difference d a2 - a1 = a3 - a2 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5 6 = 2y - 4 10 = 2y y = 5 |
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| 11603. |
What is trignometry ratio |
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Answer» Trigonometric ratios relate\xa0to the sides of a right angled triangle to\xa0its angles.specifically they are\xa0ratio of two sides of a right angled triangle and a\xa0related angle.They are used to calculate the unknown lengths or angles in a right angled triangle.\xa0e.g. let us consider a triangle ABC right Ld at B AC = hypotenuse , BC = side opposite L\xa0A , AB = adjacent side of LA\xa0then Sin A = BC / AC Cos A = AB / AC tan A = BC / AB Trigonometry ratio is the ratio of sides of the different sides of a right angled triangle |
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| 11604. |
If the equation (1+m×m) x×x + 2mcx= 0has equal root, prove that c× c =a×a (1+m×m |
| Answer» Here roots are equal,{tex}\\therefore {/tex}\xa0{tex}D = B ^ { 2 } - 4 A C = 0{/tex}Here,\xa0{tex}A = 1 + m ^ { 2 } , B = 2 m c , C = \\left( c ^ { 2 } - a ^ { 2 } \\right){/tex}{tex}\\therefore {/tex}\xa0{tex}( 2 m c ) ^ { 2 } - 4 \\left( 1 + m ^ { 2 } \\right) \\left( c ^ { 2 } - a ^ { 2 } \\right) = 0{/tex}or,\xa0{tex}4 m ^ { 2 } c ^ { 2 } - 4 \\left( 1 + m ^ { 2 } \\right) \\left( c ^ { 2 } - a ^ { 2 } \\right) = 0{/tex}or,\xa0{tex}m ^ { 2 } c ^ { 2 } - \\left( c ^ { 2 } - a ^ { 2 } + m ^ { 2 } c ^ { 2 } - m ^ { 2 } n ^ { 2 } \\right) = 0{/tex}or,\xa0{tex}m ^ { 2 } c ^ { 2 } - c ^ { 2 } + a ^ { 2 } - m ^ { 2 } c ^ { 2 } + m ^ { 2 } a ^ { 2 } = 0{/tex}or,\xa0{tex}- c ^ { 2 } + a ^ { 2 } + m ^ { 2 } a ^ { 2 } = 0{/tex}or,\xa0{tex}c ^ { 2 } = a ^ { 2 } \\left( 1 + m ^ { 2 } \\right){/tex}Hence Proved. | |
| 11605. |
If ß and alfa are zeroes of a pllynomial x2_4x+3 verify the relation between zeroes and coefficient? |
| Answer» {tex}\\alpha{/tex}\xa0and β are the zeroesThe given quadratic equation is x2—4x+3Here, a=1, b=—4, c=3By coefficient,Sum of zeroes=\xa0{tex}—b\\over a{/tex}{tex}\\alpha+β={/tex}{tex}—(-4)\\over1{/tex}=4Product of zeroes =\xa0{tex}c\\over a{/tex}{tex}\\alphaβ={3\\over1}{/tex}=3Now, x2—4x+3=0x2—3x—x+3=0x(x—3)—1(x—3)=0(x—1)(x—3)=0Either , x—1=0 or, x—3=0 The zeroes are x=1 or 3Sum of zeroes = 1+3=4Product of zeroes = 1{tex}\\times{/tex}3=3Proved | |
| 11606. |
Prove that 3+√5 is irrational ? |
| Answer» Let √3+√5 be any rational number xx=√3+√5squaring both sides x²=(√3+√5)²x²=3+5+2√15x²=8+2√15x²-8=2√15(x²-8)/2=√15as x is a rational number so x²is also a rational number, 8 and 2 are rational nos. , so √15 must also be a rational number as quotient of two rational numbers is rational but, √15 is an irrational number so we arrive at a contradiction tthis shows that our supposition was wrong so √3+√5 is not a rational number | |
| 11607. |
Rs aggarwal total q solve |
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| 11608. |
Find the coordinates of the point which divides the join of (-1,7) and (-4,3) in the ratio 2:3. |
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Answer» Let the co ordinate of the point (x,y)x={tex}mx2+nx1\\over m+n{/tex}={tex}2\\times(-4)+3\\times(-1)\\over2+3{/tex}={tex}-8-3\\over5{/tex}={tex}-11\\over5{/tex}y={tex}my2+ny1\\over m+n{/tex}={tex}2\\times3+3\\times7\\over2+3{/tex}={tex}6+21\\over5{/tex}={tex}27\\over5{/tex}The coordinate of the point is ({tex}-11\\over5{/tex},{tex}27\\over5{/tex}) (-1,7) (-4,3)√(-4+1)+(3+7)√(-3)+(4)√1 |
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| 11609. |
The first term of an AP is p and its common difference is q . Find its 10th term? |
| Answer» First term = p , common difference = q10th term = a+ (10—1)d = a+9d = p+9q | |
| 11610. |
Describe completing the square |
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| 11611. |
Find the smallest number which when divided by 28 and 32 leaves remainder 8 and 12 respectively. |
| Answer» divisor is 28 and remainder is 8. Difference is 20 (28-8)divisor is 32 and remainder is 12. Difference is 20 (32-12)You can see that the difference is 20 in both the cases.LCM of divisors - difference ,or, LCM (28, 32) - 20 = 224 - 20 = 204So, the number is 204 | |
| 11612. |
(X+Y)/XY= 2 ,(X-Y)/XY=6 solve by cross multiplication method |
| Answer» {tex}{{x + y} \\over {xy}} = 2{/tex} => {tex}{1 \\over x} + {1 \\over y} = 2{/tex} => {tex}p + q = 2{/tex} .....(i) [Let\xa0{tex}p = {1 \\over x},q = {1 \\over y}{/tex}]And {tex}{{x - y} \\over {xy}} = 6{/tex} => {tex}{1 \\over x} - {1 \\over y} = 6{/tex} => {tex}p -q = 6{/tex} .....(ii) [Let\xa0{tex}p = {1 \\over x},q = {1 \\over y}{/tex}]{tex}{p\\over {{b_1}{c_2} - {b_2}{c_1}}} = {q \\over {{c_1}{a_2} - {c_2}{a_1}}} = {{ - 1} \\over {{a_1}{b_2} - {a_2}{b_1}}}{/tex}=> {tex}{p \\over {1 \\times 6 - \\left( { - 1} \\right) \\times 2}} = {q \\over {2 \\times 1 - 6 \\times 1}} = {{ - 1} \\over {1 \\times \\left( { - 1} \\right) - 1 \\times 1}}{/tex}=> {tex}{p \\over 8} = {q \\over { - 4}} = {{ - 1} \\over { - 2}}{/tex}Taking {tex}{p \\over 8} = {1 \\over 2}{/tex} => {tex}p = 4{/tex}Taking\xa0{tex}{q \\over { - 4}} = {1 \\over 2}{/tex} => {tex}q = - 2{/tex}{tex}4 = {1 \\over x}, - 2 = {1 \\over y}{/tex} => {tex}x = {1 \\over 4},y = {{ - 1} \\over 2}{/tex} | |
| 11613. |
How many terms of the A.P:24,21,18,........multiple be taken so that their sum is 78? |
| Answer» Here a = 24, d = 21\xa0- 18\xa0= - 3, Sn\xa0= 78Sn= {tex}\\frac n2{/tex}[2a + (n - 1)d]{tex}\\therefore \\frac n2{/tex}[2(24) + (n - 1)(-3 )] = 78{tex}\\therefore \\frac n2{/tex}[48\xa0+ (n - 1)(-3 )] = 78or, n(48\xa0-3 n+ 3) = 0or, n(- 3n + 48) = 0or, n = 0 and -3n + 48\xa0= 0-3n = -48n = 48/3 = 16n = 16 | |
| 11614. |
Can anybody tell me about similar triangles? |
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| 11615. |
Nidhi saves, Rs 2 on day 1, Rs 4 on day 2 and Rs6 on day 3 how much money she saves in February 2011 |
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| 11616. |
I Nidhi saves |
| Answer» What is your question?? | |
| 11617. |
If a cos^3 theta +3a cos theta sin ^2 theta =m |
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| 11618. |
Sample paper for 2017 2018 provide karaiye |
| Answer» Check Sample papers here :\xa0https://mycbseguide.com/cbse-sample-papers.html | |
| 11619. |
Elimination method |
| Answer» The elimination method of solving systems of equations is also called the addition method. To solve a system of equations by elimination we transform the system such that one variable "cancels out". | |
| 11620. |
Why we add 1 in converse of theorem in triangle |
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| 11621. |
Solution of Ch-Triangle Ex-6.3 Q14 is wrong in your ncert solution.Correct this solution |
| Answer» Given :∆ABC and ∆PQR, AD and PM are mediansBD= BC/2 and QR = QR/2So, AB/PQ = AC/PR=AD/PMTo prove:∆abc ~∆pqrConstruction:ad extended upto e and PM extended upto L.join be,CE,ql and elAd=de and PM=mlBD=DC and QM=mrTherefore,abce and pqrl are ||gramAb=CE and AC=be PQ= elPr=qlIn ∆abe and ∆pqlAb/PQ=BC/QR=ad/pmAb/PQ=BC/2×2/QR=2ad/2pmAb/PQ=bd\\qm=ae/pl∆abe~∆pql by SSS ruleangle bae=angle qpl (equation first)Similarly ∆ace~∆prlTherefore angle cae=RPL (equation second)Adding equation first and secondAngle bae + cae=angle qpl+RPLAngle bac = Apr (equation third)In ∆abc and ∆pqrAngle bac =qpr from equation thirdAB/PQ=BC/QR (given)Therefore ∆ABC ~∆PQR (SAS)Proved | |
| 11622. |
Ncert ex 12.2 Q 6 with explanation anyone please help |
| Answer» Here, Area of equilateral triangle =\xa0{tex}{{\\sqrt 3 } \\over 4}{\\left( {{\\rm{Side}}} \\right)^2}{/tex}=\xa0{tex}{{\\sqrt 3 } \\over 4}{\\left( {{\\rm{15}}} \\right)^2}{/tex}\xa0= 97.425 sq. cmArea of circle =\xa0{tex}\\pi {r^2} = \\pi {\\left( {15} \\right)^2}{/tex}\xa0= 706.5 sq. cmArea of minor sector =\xa0{tex}{1 \\over 6} \\times 706.5{/tex}\xa0= 117.75 sq. cmTherefore, Area of minor segment = 117.75 - 97.425 = 20.325 sq. cmAnd, Area of major segment = 706.5 - 20.325 = 686.175 sq. cm | |
| 11623. |
What is bpt |
| Answer» Basic Proportionality Theorem (Thales theorem): If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio. | |
| 11624. |
12x+5y=35 |
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| 11625. |
What is pie |
| Answer» The correct spelling is "Pi".The number π is a mathematical constant, the ratio of a circle\'s circumference to its diameter, commonly approximated as 3.14. | |
| 11626. |
If tan A+ cot A= 2 then find the value of tan2 A + cot2 A |
| Answer» I don\'t know | |
| 11627. |
SinA+cosA |
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| 11628. |
Plese tell me important question for10 class in all chapters |
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| 11629. |
Half yearly exam\'s marks will add in final result or not plz tell |
| Answer» No | |
| 11630. |
Ex 12.2 Q no.7 plz tell solution with explanation plz help |
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| 11631. |
If x=3 ,y=9-3, find (x^2+y^2)xy+3xy |
| Answer» 774 | |
| 11632. |
Who made the subject Mathematics.please tell his name and adsress |
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Answer» Address- heaven. You will have to die. Maths in India emerged in 1200 BCE. It was brought by Aryabhatta, Brahmagupta, Mahavira , Bhaskara II , Madhava of Sangamagrama and Nilkantha Somayaji. |
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| 11633. |
The least positive integer n for which underroot n+1-UNDERROOT n-1 |
| Answer» Let us assume that there is a positive integer n for {tex}\\sqrt{n-1}+\\sqrt{n+1}{/tex}which is rational and equal to {tex}\\frac pq{/tex}, where p and q are positive integers and (q\xa0{tex}\\neq{/tex}\xa00).{tex}\\sqrt { n - 1 } + \\sqrt { n + 1 } = \\frac { p } { q }{/tex}......(i)or,\xa0{tex}\\frac { q } { p } = \\frac { 1 } { \\sqrt { n - 1 } + \\sqrt { n + 1 } }{/tex}on multiplication of numerator and denominator by\xa0{tex}\\sqrt{n-1}-\\sqrt{n+1}{/tex}\xa0we get{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( \\sqrt { n - 1 } + \\sqrt { n + 1 } ) ( \\sqrt { n - 1 } - \\sqrt { n + 1 } ) }{/tex}{tex}= \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { ( n - 1 ) - ( n + 1 ) } = \\frac { \\sqrt { n - 1 } - \\sqrt { n + 1 } } { - 2 }{/tex}or,\xa0{tex}\\sqrt { n + 1 } - \\sqrt { n - 1 } = \\frac { 2 q } { p }{/tex} ........(ii)On adding (i) and (ii), we get{tex}2 \\sqrt { n + 1 } = \\frac { p } { q } + \\frac { 2 q } { p } = \\frac { p ^ { 2 } + 2 q ^ { 2 } } { p q }{/tex}{tex}\\sqrt{n+1}\\;=\\frac{p^2+2q^2}{2pq}{/tex}...............(iii)From (i) and (ii),{tex}\\style{font-family:Arial}{\\sqrt{n-1}\\;=\\frac{p^2-2q^2}{2pq}}{/tex}........(iv)In RHS of (iii) and (iv)\xa0{tex}\\frac{p^2+2q^2}{2pq}\\;and\\;\\frac{\\displaystyle p^2-2q^2}{\\displaystyle2pq}\\;are\\;rational\\;number\\;because\\;p\\;and\\;q\\;are\\;positive\\;integers{/tex}But it is possible only when (n + 1) and (n - 1) both are perfect squares.Now n+1-(n-1)=n+1-n+1=2Hence they differ by 2 and two perfect squares never differ by 2.So both (n + 1) and (n -1 ) cannot be perfect squares. Hence there is no positive integer n for which\xa0{tex}\\style{font-family:Arial}{\\sqrt{n-1\\;}+\\sqrt{n+1}}{/tex} is rational | |
| 11634. |
Triangle |
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| 11635. |
How to prove that root 2 is irrational |
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Answer» Let us assume that |
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| 11636. |
Sin 45 |
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| 11637. |
What will be the roots of this quadratic equation x2 -3x =0 |
| Answer» x2-3x=0x(x-3)=0 [by taking x common]x=0 or x-3=0x=0 x=3So roots of the given quadratic equation are x= 0 and x= 3. | |
| 11638. |
Show that (A-B)^2,(A^2+B^2),(A+B)^2 is an AP |
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Answer» To Prove (A-B)2\xa0,A2+B2 and (A+B)2\xa0is in an AP, We need to show that the difference between the terms are equal.Let a1= (A-B)2 a2=A2\xa0+B2 and a3= (A+B)2Now a2-a1\xa0=\xa0A2\xa0+B2\xa0-\xa0(A-B)2\xa0 and a3\xa0- a2\xa0=(A+B)2\xa0- (A2\xa0+ B2) = A2+B2\xa0-(A2\xa0-2AB+B2) =A2+2AB+B2\xa0-A2\xa0- B2 = A2\xa0+B2\xa0- A2\xa0+2AB - B2 = 2AB = 2ABClearly a2-a1\xa0= a3\xa0- a2\xa0= 2ABSo given terms are in AP with common difference d= 2AB 121221112111111111111111111111111 |
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| 11639. |
Find HCF of 96and 404 by prime factors method. |
| Answer» Prime factor of 96 is 2,2,2,2,2,3Prime factor of 2,2,101HCF=2,2 | |
| 11640. |
4+5 |
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| 11641. |
If sinthitha- costhitha+1/sinhitha-costhitha+1 = tanthitha- cosecthitha |
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| 11642. |
What is EUCLID LEMMA |
| Answer» It is a division where,we can write in the form of a=bq+r where,0< or= to r | |
| 11643. |
Prove that root5irrational |
| Answer» Let us assume that √5 be rational √5= p/q. ( where p and q are integers ,q is not equal to 0, p,q donot have common factor). 1.. √5square= p/q square . 5=p square/q square . 5 is a factor of p square 5 is a factor of p P is multiple of 5 Let p be 5K 5q square=5K whole square . 5q square =25K square. q square=5 k square. 5 is a factor of q square. 5 s factor of q...........3. From 2....3...... p and q have a common factor......4. | |
| 11644. |
Find the pair of natural numbers whose lcm is 78 and greatest divisor is 13 |
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Answer» LCM = 78GCD = 13 (greatest common divisor)LCM X GCD = 78 X 13 = 1014or 2 X 3 X 13 X 13since 13 is GCD , therefore 13 should be multiple of both numbers\xa0so the numbers are 2 x 13 and 3 x 13 or 26 , 39\xa0 39 and 26 |
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| 11645. |
If a+b+c=1 a^2+b^2+c^2=2 a^3+b^3+c^3=3Then find the value of , a^4+b^4+c^4. |
| Answer» \xa0{tex}{\\left( {a + b + c} \\right)^3} = {a^3} + {b^3} + {c^3} + \\left( {a + b + c} \\right)\\left( {ab + bc + ca} \\right){/tex}=> {tex}{\\left( 1 \\right)^3} = 3 + \\left( 1 \\right)\\left( {ab + bc + ca} \\right){/tex}=> {tex}ab + bc + ca = - 2{/tex}Squaring both sides, we get{tex}{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + 2abc\\left( {{a^2} + {b^2} + {c^2}} \\right) = 4{/tex}=> {tex}{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + 4abc = 4{/tex}=> {tex}{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} = 4 - 4abc{/tex} ..............(i){tex}{a^2} + {b^2} + {c^2} = 2{/tex}Squaring both sides, we get=> {tex}{a^4} + {b^4} + {c^4} + 2\\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \\right) = 4{/tex}=> {tex}{a^4} + {b^4} + {c^4} + 2\\left( {4 - 4abc} \\right) = 4{/tex} [From eq.(i)]=> {tex}{a^4} + {b^4} + {c^4} + 8 - 8abc = 4{/tex}=> {tex}{a^4} + {b^4} + {c^4} = - 4 + 8abc{/tex} ..................(ii)Now, using one result,\xa0{tex}6abc = {\\left( {a + b + c} \\right)^3} + 2\\left( {{a^3} + {b^3} + {c^3}} \\right) - 3\\left( {a + b + c} \\right)\\left( {{a^2} + {b^2} + {c^2}} \\right){/tex}=> {tex}6abc = {\\left( 1 \\right)^3} + 2\\left( 3 \\right) - 3\\left( 1 \\right)\\left( 2 \\right){/tex}=> {tex}abc = {1 \\over 6}{/tex}Putting this value in eq.(ii), we get{tex}{a^4} + {b^4} + {c^4} = - 4 + 8 \\times {1 \\over 6}{/tex}=> {tex}{a^4} + {b^4} + {c^4} = {{ - 8} \\over 3}{/tex}\xa0 | |
| 11646. |
What is the hcf of smallest composite number & smallest prime number ? |
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Answer» Ans is 2 Got the answer it\'s 4&2 2 |
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| 11647. |
How we can do solution of quadratic equation by method of completing the square |
| Answer» {tex}x^2 +\xa06x - 16 = 0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2 + 6x = 16{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}x^2 + 6x + 9 = 16 + 9{/tex} [Adding on both sides square of coefficient of x, i.e. ({tex}\\frac{6}{2}{/tex})2]{tex}\\Rightarrow{/tex}\xa0{tex}(x + 3)^2 = 25{/tex}{tex}\\Rightarrow{/tex}\xa0x + 3 =\xa0{tex}\\pm{/tex}{tex}\\sqrt{25}{/tex}{tex}\\Rightarrow{/tex}\xa0x + 3 =5 or x + 3 = -5{tex}\\Rightarrow{/tex}\xa0x = 2 or x = -8 | |
| 11648. |
what are the advantage s of having four chamberd of heart |
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Answer» Advantages: oxygen rich blood and corbondioxide rich blood do not mix together. :It is easy to pump both the blood simntinously Having a four chambered heart helps in easy circulation and also separates oxygenated blood from deoxygenated blood |
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| 11649. |
4x square-2(k+1)x+k+4? |
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| 11650. |
is only ncert not sufficient for trignometry and some application of trignometry??. |
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Answer» Use full marks Yes not sufficient. Use RS Agarwal last year edition have harder questions than new. Use r.d.sharma for practise |
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