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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.
| 39601. |
AM is a median of a triangle ABC is AB+BC>2AM? |
| Answer» In {tex}\\triangle{/tex}ABMAB + BM > AM....[Sum of the lengths of any two sides of a triangle is greater than the length of the third side].....(1)In {tex}\\triangle{/tex}ACMCA + CM > AM....[Sum of the lenghts of any two sides of a triangle is greater than the length of the third side].....(2)Sum (1) and (2)(AB + BM) + (CA + CM) > AM + AM{tex}\\therefore{/tex}\xa0AB + (BM + CM) + CA > 2AM{tex}\\therefore{/tex}\xa0AB + BC + CA > 2AM | |
| 39602. |
Tan A+CotA=2 than find the valu of Tan squarA- cot squareA |
| Answer» Given, tan A + cot A = 2We need to square\xa0both sides,\xa0(tan A + cot A)2 = (2)2{tex}\\Rightarrow{/tex}tan2A + cot2A + 2 tan A. cotA = 4\xa0{tex}[(a+b)^2=a^2+b^2+2ab]{/tex}{tex}\\Rightarrow{/tex}tan2A + cot2A + 2 tan A{tex}\\times \\frac { 1 } { \\tan A }{/tex}= 4{tex}\\Rightarrow{/tex}tan2A + cot2A + 2 = 4{tex}\\Rightarrow{/tex}tan2A + cot2A = 4 - 2 =\xa02Therefore the value of\xa0tan2 A + cot2 A = 2 | |
| 39603. |
maths ex7.1solution |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 39604. |
Find the value of k if 8k+4,6k-2 and 2k-7 are in ap |
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Answer» Difference=d,d=d,6k-2-(8k+4)=2k-7-(6k-2),6k-2-8k-4=2k-7-6k+2,-2k-6=-4k-5,2k=11,k=11/2 12k-4=8k+4+2k-712k-10k =-3+42k= 1k=1/2 |
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| 39605. |
CosA 0.6,show that(5sinA-3tanA)=0 |
| Answer» Given,\xa0cos{tex}\\theta{/tex}\xa0= 0.6\xa0{tex}= \\frac { 6 } { 10 } = \\frac { 3 } { 5 }{/tex}Let us draw a triangle ABC in which\xa0{tex}\\angle{/tex}B =\xa090°.Let\xa0{tex}\\angle{/tex}A =\xa0{tex}\\theta{/tex}°.Then,\xa0{tex}\\cos \\theta = \\frac { A B } { A C } = \\frac { 3 } { 5 }{/tex}Let AB = 3k and AC = 5k, where k is positive.By Pythagoras\' theorem, we haveAC2\xa0= AB2 + BC2{tex}\\Rightarrow{/tex}BC2\xa0= AC2\xa0- AB2\xa0= (5k)2\xa0- (3k)2\xa0= 25k2\xa0- 9k2\xa0= 16k2{tex}\\Rightarrow \\quad B C = \\sqrt { 16 k ^ { 2 } } = 4 k{/tex}{tex}\\sin \\theta = \\frac { A B } { A C } = \\frac { 4 k } { 5 k } = \\frac { 4 } { 5 }{/tex}{tex}\\cos \\theta = \\frac { 3 } { 5 }{/tex}{tex}\\tan \\theta = \\frac { \\sin \\theta } { \\cos \\theta } = \\left( \\frac { 4 } { 5 } \\times \\frac { 5 } { 3 } \\right) = \\frac { 4 } { 3 }{/tex}{tex}\\Rightarrow ( 5 \\sin \\theta - 3 \\tan \\theta ) = \\left( 5 \\times \\frac { 4 } { 5 } - 3 \\times \\frac { 4 } { 3 } \\right) = 0{/tex}Hence,\xa0(5sin{tex}\\theta{/tex}\xa0- 3 tan{tex}\\theta{/tex}) = 0. | |
| 39606. |
A:B=2:6 B:C=5:7 and C:D=1:9 find A:B:C:D |
| Answer» | |
| 39607. |
Prove that:tan theta - 1 + sec theta / tan theta + 1-sec theta = 1 / sec theta - tan theta |
| Answer» L.H.S={tex}\\frac { \\tan \\theta + \\sec \\theta - 1 } { \\tan \\theta - \\sec \\theta + 1 }{/tex}{tex}= \\frac { ( \\tan \\theta + \\sec \\theta ) - \\left( \\sec ^ { 2 } \\theta - \\tan ^ { 2 } \\theta \\right) } { \\tan \\theta - \\sec \\theta + 1 }{/tex}\xa0{tex}[\\because sec^2\\theta-tan^2\\theta=1]{/tex}{tex}= \\frac { ( \\tan \\theta + \\sec \\theta ) - ( \\sec \\theta - \\tan \\theta ) ( \\sec \\theta + \\tan \\theta ) } { \\tan \\theta - \\sec \\theta + 1 }{/tex}{tex}= \\frac { ( \\tan \\theta + \\sec \\theta ) [ 1 - \\sec \\theta + \\tan \\theta ] } { \\tan \\theta - \\sec \\theta + 1 }{/tex}=\xa0{tex}tan\\theta+sec\\theta{/tex}= R.H.S.Hence Proved. | |
| 39608. |
Can i take 90 percent without practicing rd sharma in maths |
| Answer» Yes but hard work is needed | |
| 39609. |
What is the answer of excercise 3.7 |
| Answer» It is given in this app | |
| 39610. |
Find the root of the following quadratic equation by factorisation:1. 2 X square + x - 6 = 0 |
| Answer» Split the middle term in 4x-3x | |
| 39611. |
integration under root 3 X square + 4 x + 1 DX |
| Answer» | |
| 39612. |
If nth term of an AP is 5-11n . find common difference |
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Answer» -11 is the right answer. an= 5-11na¹=5-11(1)=-6a²=5-11(2)=-17d=a²-a¹= -17-(-6)=-11Common difference is -11 23 |
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| 39613. |
Show that n square -1 is divisible by 8 |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 39614. |
Last 5 cbse exam paper |
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Answer» Search on google You can use Google |
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| 39615. |
how can we calculate ph value of any solution ? |
| Answer» By ph paper or ph solution | |
| 39616. |
For an AP show that Tp+2q equal 2Tp+q |
| Answer» Let the first term be a and the common difference be d.we know,\xa0{tex}a_n= a+(n-1)d{/tex}{tex}a_p=a+(p-1)d{/tex}{tex}a_{p+2q}=a+(p+2q-1)d{/tex}\xa0{tex}\\therefore a_p+a_{p+2q}=a + (p - 1)d + a + (p + 2 q -1)d{/tex}{tex}= a + pd - d + a + pd + 2qd - d{/tex}{tex}= 2a + 2pd + 2qd - 2d{/tex}{tex}= 2[a + (p + q - 1) d]{/tex} ............(i){tex}2a_{p+q}=2[a + (p + q - 1 ) d ]{/tex}\xa0..........(ii)From (i) and (ii), we getap + ap + 2q\xa0= 2ap+q | |
| 39617. |
If two points P(x,y) is equidistant from the points A(a+b,a-b) and B(a-b,a+b).Prove that bx=ay |
| Answer» |PQ| = |PR{tex}\\begin{aligned} \\sqrt { [ x - ( a + b ) ] ^ { 2 } + [ y - ( b - a ) ] ^ { 2 } } = \\sqrt { [ x - ( a - b ) ] ^ { 2 } + [ y - ( b + a ) ] ^ { 2 } } \\end{aligned}{/tex}Squaring, we get[x - (a + b)]2 + [y - (b\xa0- a)]2\xa0= [x - (a - b)]2 + [y - (a + b)]2or, [x - (a + b)]2 - [x - a + b]2\xa0= (y - a - b)2 - (y - b + a)2or, (x - a - b + x - a + b) ( x - a - b - x + a - b)= (y - a - b + y - b + a)(y - a - b - y + b - a)or, (2x - 2a) (- 2b) = (2y - 2b) (- 2a)or, (x - a)b = (y - b)aor, bx = ay.Hence Proved. | |
| 39618. |
3d+1=5 |
| Answer» 4/3 | |
| 39619. |
What is the degree of zero |
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Answer» Not defined but may be in a negative form like -1 There may be degree of zero but till now it is not discovered There is no degree of o May be 0 |
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| 39620. |
Find the value of A if 2sin 2A=√3 |
| Answer» | |
| 39621. |
2+2×98÷36 |
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Answer» 10.88888889 5.5 67/9 |
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| 39622. |
√2 |
| Answer» What is this..... what we do | |
| 39623. |
For what value of K will the consecutive terms 2k+1,3k+3,and 5k-1 form and A.P |
| Answer» K = 6 | |
| 39624. |
Find the distance between the following pairs ofpoints |
| Answer» Where r the pair of points | |
| 39625. |
What is mid point formula |
| Answer» X2+x1÷2 , y2+y1÷2 | |
| 39626. |
3(r-7)=4(r-8) |
| Answer» | |
| 39627. |
Syllabus class 10 ,2018-19 |
| Answer» Check syllabus here :\xa0https://mycbseguide.com/cbse-syllabus.html | |
| 39628. |
Theorem 6.3: If in two triangles, corresponding |
| Answer» | |
| 39629. |
If567+654/64 |
| Answer» | |
| 39630. |
2x+6y |
| Answer» What is 2x+6y | |
| 39631. |
Cota+coseca-1/cota-coseca+1= 1+cosa/sina |
| Answer» | |
| 39632. |
Show that any square of an odd positive integer is of the form 5q ,5q+1,5q+4 for some integer q |
| Answer» Any integer can be written in the form 5m , 5m+1, 5m+2 (where m is any integer)(5m)² = 25m² = 5 × 5m² = 5q (where q = 5m²)(5m+1)² = (5m)² + 2 ×5m × 1 + 1² [by using identity- (a+b)²= (a²+ 2ab +b² )] = 25m² + 10 m + 1 = 5 (5m²+ 2m) +1 = 5q +1 ( where q= 5m²+2m)(5m+2)²= (5m)² + 2× 5m × 2 +2² [by using identity- (a+b)²= (a²+ 2ab +b² )] = 25m² + 20 m +4 = 5 (5m²+4m ) + 4 = 5q+4 (where q= 5m²+4m)Hence proved | |
| 39633. |
Another name of an |
| Answer» | |
| 39634. |
Find the value of k if x2+2x+k is a factor of 2x4+14x2+5x+6Also find all the zeroes of polynomial |
| Answer» If g(x) = x2 + 2x + k is a factor of f(x) = 2x4 + x3 - 14x2 + 5x + 6, then remainder is zero when f(x) is divided by g(x).Let quotient = Q and remainder = RLet us now divide f(x) by g(x).R = x(7k + 21) + (2k2 + 8k + 6) -------(1) and Q = 2x2 - 3x - 2(k + 4).------------(2)Now, R = 0.{tex}\\Rightarrow{/tex}\xa0x (7k + 21) + 2 (k2 + 4k + 3) = 0\xa0{tex}\\Rightarrow{/tex}\xa07x (k + 3) + 2 (k+1)(k+3) = 0{tex}\\Rightarrow{/tex}\xa0(k+3) [7x + 2(k+1)] = 0{tex}\\Rightarrow{/tex}\xa0k + 3 = 0{tex}\\Rightarrow{/tex}\xa0k = -3Thus, polynomial f(x) can be written as,2x4 + x3 - 14x2 + 5x + 6 = (x2 + 2x + k) [2x2 - 3x - 2(k + 4)] = (x2 + 2x - 3) (2x2 - 3x - 2)Zeros of\xa0x2 + 2x - 3 are,x2 + 2x - 3 = 0{tex}\\Rightarrow{/tex}\xa0(x + 3) (x - 1) = 0{tex}\\Rightarrow{/tex}\xa0x = -3 or x = 1Zeros of\xa0(2x2 - 3x - 2) are,2x2 - 3x - 2 = 0{tex}\\Rightarrow{/tex}\xa02x2 - 4x + x - 2 = 0{tex}\\Rightarrow{/tex}\xa02x(x - 2) + 1(x - 2) = 0{tex}\\Rightarrow{/tex}\xa0(x - 2)(2x + 1) = 0x = 2 or x = -{tex}\\frac12{/tex}Thus, the zeros of f(x) are: -3 ,1, 2 and\xa0-{tex}\\frac12{/tex} | |
| 39635. |
Chapter 14 formual |
| Answer» | |
| 39636. |
if the diagonals of a quadrilateral divide each other proportinally then it is a trapezium |
| Answer» No | |
| 39637. |
Ex4 all formula |
| Answer» | |
| 39638. |
A thef |
| Answer» | |
| 39639. |
Find the middle term of the AP 6;13;20:::; 216 |
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Answer» 107.5 31 |
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| 39640. |
3+8 |
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Answer» 2q 11 |
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| 39641. |
Chapter 8 all formula |
| Answer» | |
| 39642. |
If sin theta=cos theta find the value of theta |
| Answer» 45degree | |
| 39643. |
{2-3(2-3)Δ}Δ |
| Answer» | |
| 39644. |
Blueprint board exam 10class 2018-2019 |
| Answer» | |
| 39645. |
secA=13/5 ,show that 2sinA-3cosA/4sinA-9cosA=3 |
| Answer» Given that SecA=13/5. Then we know that Sec=hypotenius/Base, I.e,13/5Then we find perpendicular =12Thus, 2 sinA - 3cosA /4sinA -9cosA =3 {(2 ✖ 12/13 - 3 ✖ 5/13)/(4 ✖ 12/13 - 9 ✖ 5/13)} = 3or. {24 -15/48-45}=3 or. { 9/3}= 3Or. 3=3 Hence LHS=RHS | |
| 39646. |
Find the value of k for which 2x2 +kx+3 =0 has two equal and real roots. |
| Answer» Since the eqaution has two equal roots, then b^2 - 4ac = 0== k^2 - 4×2×3=0 == k^2 - 24 == k^2=24 == k=+-√24 == k= +-2√6 | |
| 39647. |
What is BPT therom? |
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Answer» If a line is drawn parallel to one side of a triangle then it Intersects the other two sides of the triangle in the same ratio basic proportional theorem Basic proportionality theouram |
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| 39648. |
Is (a+b)2 formlue |
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Answer» a2+2ab+b2 a2+b2+2ab a2+b2+2ab Yes a plus b whole sq. is a formulae |
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| 39649. |
Rs agrawal ka questions and answers chahiye |
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Answer» Target nhi hai kya R.D. nhi hai kya? |
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| 39650. |
Prof That Exterional angel of triangle sum of opposite angel |
| Answer» | |