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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.
| 16501. |
x/a+y/b=2ax-by=a^2-b^2 |
| Answer» x=a;y=b | |
| 16502. |
How to find modal class in statistics |
| Answer» Highest frequency is modal class | |
| 16503. |
What is upstream and downstream |
| Answer» Upstream means opposite direction of stream and Downstream means along with stream.? | |
| 16504. |
How to solve subtitution method in liniear equation in to variable |
| Answer» In substitution method . initially you saw that there are two equation in the form of x and y or any other variable , firstly you should take one equation and find the vlue of one variable for example---- suppose first equation will be ( x+y = 4) and another one is (y+ 2x= 5) now take equation which is (x+y= 4 ) == ( x= 4-y ) and put this in another equation which is ( y+2x= 5) Then solve the value for x and y . | |
| 16505. |
Find the zeroes of quadratic equations :25x2+25x+4 |
| Answer» Your first term is wrong | |
| 16506. |
Ex-14.1 Qno.3 plz explain how can we cancel 18 |
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Answer» From the given data mean is(a) 18 therefore (x)=a+sigma. Fidi/sigmafi=18=18+2f-40/44+f2f-40/44+f=02f-40=02f=40F=20Hence missing frequency is 20? Ncert or ......???? |
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| 16507. |
Find the sum of first 10 multiples of 3 |
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Answer» Sorry the formula is Sn=n÷2(2a+(n-1)×d) 165.... The no.of terms is 10 so, n=10The first term is 3 so, a=3The the multiples of 3 means the common difference is also 3 so,d=3Then substitute the value of (a,n,d)in the formula. Sn=n÷2(2n+(n-1)+d) 165 |
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| 16508. |
Write the discriminant of the quadratic equation (x+5) whole square = 2(5x-3) |
| Answer» Discriminant =-125 | |
| 16509. |
25x(x+1)=-4 find quadratic equations by polynomial |
| Answer» -1/5and-4/5 | |
| 16510. |
Find the first term, common difference and next term of each Q. 9, 15, 21, 27....... |
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Answer» A=9 d=6 & a(n)=55th=9+(5-1)6=9+24=333 First term=a=9Common difference =d =15-9=65th term=a+(n-1)d =9+(5-1)6 =9+24 =33 |
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| 16511. |
If tan alpha =5/12 find the value of sec alpha |
| Answer» According to the question,\xa0tan\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\frac{5}{12}{/tex}We know that, sec\xa0{tex}\\alpha{/tex}\xa0=\xa0{tex}\\sqrt{1 + tan^2 \\alpha}{/tex}=\xa0{tex}\\sqrt{1 + \\frac{25}{144}}{/tex}\xa0={tex}\\sqrt{ \\frac{144+25}{144}}{/tex}={tex}\\sqrt{ \\frac{169}{144}}{/tex}=\xa0{tex}\\frac{13}{12}{/tex} | |
| 16512. |
What is the value of triqngle |
| Answer» Which triangle right, scalane ,isosceles? ?? | |
| 16513. |
T*T-15 polynomial |
| Answer» T*T-15=0Now,Put, t=0T-15=0So,T=15Therefore,zeroes are (0,15) | |
| 16514. |
The perimeter of rhombus abcd is 68 . If ac = 30 find bd |
| Answer» You have big tits and i got big **** so suck my **** | |
| 16515. |
Triangle abc ~triangle pqr for the correspondence . If ab +BC = 12 ,PQ + qr= 15 and ac=8 find pr |
| Answer» Which school in which state | |
| 16516. |
□ abcd is a rectangle .if ab=7.5 and ac = 19.5 find bc |
| Answer» You have big boobs | |
| 16517. |
What is mean by circle |
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Answer» C(r,c) Which is closed curved figure A circle is a collection of points which are equidistant from a point A circle is a simple Closed shape It is a Set Of all point In a plane .That are at agiven Distance for a givan Point At the centre equvalentry is called a circle A circle is a geometrical figure whose all radius is equal from centre |
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| 16518. |
Write wheter every integer can be of form 4q +2 , where q is an integer. Justify |
| Answer» | |
| 16519. |
1377 |
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Answer» What\'s wrong with this number? What 1377 |
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| 16520. |
RD SHARMA ex3.4 solutions |
| Answer» U can Search on internet also ? | |
| 16521. |
PA÷AQ=PB÷BR =3 if area of triangle PQR=32 then find area of quaderlaetral AQBR |
| Answer» | |
| 16522. |
show that any positive odd integer is of the form (4m+1) or (4m+3) where m is some integer. |
| Answer» Let a be the positive integer.And, b = 4 .Then by Euclid\'s division lemma,We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .°•° Then, possible values of r is 0, 1, 2, 3 .Taking r = 0 .a = 4q .Taking r = 1 .a = 4q + 1 .Taking r = 2a = 4q + 2 .Taking r = 3 .a = 4q + 3 .But a is an odd positive integer, so a can\'t be 4q , or 4q + 2 [ As these are even ] .•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .Hence , it is solved\xa0 | |
| 16523. |
Find the point on x-axis which is equidistant from points (-1,0) and (5,0) |
| Answer» Let A(x,o) be any point on the X-axis , which is equidistant from points (-1,0) and (5,0).{tex}\\Rightarrow (x+1)^2=(x-5)^2{/tex}⇒ x2 + 2x + 1 = x2 - 10x + 25⇒ 2x + 1 = -10x + 25⇒ 2x + 10x = 25 - 1⇒ 12x = 24⇒ x = 24/12{tex}\\Rightarrow x= 2{/tex}Therefore , Required point is (2,0). | |
| 16524. |
Is there any easy method to make equations ?? |
| Answer» Oops . | |
| 16525. |
Solution of this questionFind two consecutive positive integers sum of whose square is 365. |
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Answer» 14 Let be the two consecutive +ve integers are x and ( x+1)According to the question — x² + (x+1)² = 365 x² + x² + 2x + 1 = 365 [using (a+b)² = a² +2ab+ b²]2x² +2x = 365 - 12x² + 2x - 364 = 0Dividing above equation by 2 ,we get —x² + x - 182 = 0 x² + (14-13)x - 182 = 0x² + 14x - 13x - 182 = 0x(x+14) -13(x+14) = 0(x + 14)(x - 13) = 0So , (x = -14) & (x = 13) Since , positive cosecutive integer can never be negative .Hence, two consecutive positive integers are : x = 13 & (x + 1) = 14 Ans. Ans. |
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| 16526. |
Hcf of 12576 and 4052 |
| Answer» HCF of 12576 and 4052 is 4 | |
| 16527. |
x÷2+2y÷3=-1 and x-y÷3=3Solve in elimination method |
| Answer» x/2+2y/3=-1 Take lcm and solve3x+4y=-6 mark it as 1We will get3x-y=9 after solving equation 2From 1 and 2,3x+4y=-63x-y=9(-) (+) (-)5y=-15y=-3x=2 | |
| 16528. |
Find the sum of the first 25 terms of an AP whose nth term is given by an=7-3n |
| Answer» Given, an\xa0= 7 - 3nPut\xa0n = 1, a1\xa0= 7\xa0- 3\xa0{tex}\\times{/tex}\xa01 = 7\xa0- 3 = 4Put\xa0n = 2, a2\xa0= 7\xa0- 3\xa0{tex}\\times{/tex}\xa02 = 7\xa0- 6 = 1Common difference(d) = 1 - 4\xa0=\xa0-3Sn\xa0=\xa0{tex}\\frac { n } { 2 } [ 2 a + ( n - 1 ) d ]{/tex}S25\xa0=\xa0{tex} \\frac { 25 } { 2 } [ 2 \\times 4 + ( 25 - 1 ) ( - 3 ) ]{/tex}{tex}= \\frac { 25 } { 2 } [ 8 - 72 ]{/tex}{tex}= \\frac { 25 } { 2 } \\times - 64{/tex}= -800 | |
| 16529. |
Root 5 +root 5 |
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Answer» 2√5 2√5 is the correct answer 2 root five 2√5 2√5 5 2√5 |
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| 16530. |
6th maths playing with numbers |
| Answer» | |
| 16531. |
Waht is the place value of 4 in 874569036 in the |
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Answer» 40,00,000 4000000 400000 Common hai 40,00,000 40,00,000 is the place value 4 |
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| 16532. |
Who\'s chapter is very important fo Board exam |
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Answer» All subjects and chapters are Important in board exam Most important chapter is tenomatery All course are important in board exams and also improves your knowledge which is given in syallbus |
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| 16533. |
How to questions via quadratic equations? |
| Answer» X square + x + 4 = 0 | |
| 16534. |
Ncert chapter 5 all formulas |
| Answer» Check revision notes to check formulae :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 16535. |
Sin2 kakaiawn |
| Answer» | |
| 16536. |
One zero of a polynomial 3x*x -8x+12k+1 is 7 times the other find the value of k. |
| Answer» math is not easy to solve in mobile | |
| 16537. |
It\'s really a true news that maths have 2 papers in 2020 |
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Answer» yes it\'s true news that maths having 2 paper that is BASIC and STANDARD Yeah !! Basic and Standard |
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| 16538. |
6 x squre _ 3_7 z |
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Answer» Me too ? i cannot understand |
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| 16539. |
Find k |
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Answer» Whats you mean this question is wrong wrong question |
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| 16540. |
Formulae to find \'d\' of an A.P |
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Answer» d= b^2-4ac is the formula of d Common difference d =a2- a1 d=b^2-4ac d=a2-a1 |
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| 16541. |
X cube + Y cube= |
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Answer» Sorry you all are wrong (x+y)(x^2+y^2-xy) (x+y)^3 -3xy (x+y) Sorry it is (x+y)^3+3xy(x+y) (x+y)^3 +3xy(x-y) |
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| 16542. |
If -1 and 2 are the two zeros of the polynomial 2x cube-x square-5x-2 , its third zero is _________ |
| Answer» Given polynomial is p(x) = 2x3\xa0- x2- 5x - 2\xa0and -1 and 2 are zeroes of polynomial.{tex}\\therefore{/tex}\xa0{x - (-1)} (x - 2)= ( x + 1) (x - 2) = x2 - 2x + x - 2 = x2- x - 2 is a factor of p(x)For other zeroes, 2x\xa0+ 1 = 0{tex}\\Rightarrow x = \\frac { - 1 } { 2 }{/tex}{tex}\\therefore{/tex}\xa0Other zero = {tex}\\frac { - 1 } { 2 }{/tex} | |
| 16543. |
If the product of zeros of the polynomial ax2-6x-6 is 4, then the value of a is ---------- |
| Answer» According to the question,we have to find the value of a such that the product of the zeros of the polynomial (ax2\xa0-\xa06x -\xa06) is 4.Let {tex} \\alpha{/tex}\xa0and {tex}\\beta{/tex}\xa0be the zeros of the polynomial (ax2\xa0- 6x - 6)Then,\xa0{tex}\\alpha{/tex}{tex} \\beta{/tex}\xa0=\xa0{tex} \\frac { \\text { constant term } } { \\text { coefficient of } x ^ { 2 } } = \\frac { - 6 } { a }{/tex}But,\xa0{tex} \\alpha{/tex}{tex} \\beta{/tex}\xa0= 4 (given).{tex} \\therefore \\quad \\frac { - 6 } { a } = 4 \\Rightarrow 4 a = - 6 \\Rightarrow a = \\frac { - 6 } { 4 } = \\frac { - 3 } { 2 }{/tex}Hence, a =\xa0{tex} \\frac { - 3 } { 2 }{/tex} | |
| 16544. |
√secQ-1+√secQ+1=2secQ |
| Answer» | |
| 16545. |
If secQ+tanQ=p find the value if cosQ |
| Answer» | |
| 16546. |
If 19th term is equal to three times it 6th term and 9 th term is 19 find ao |
| Answer» Let the first term of the A.P. be \'a\'.and the common difference be \'d\'.19th term of the A.P., t19 = a + (19 - 1)d = a + 18d6th term of the A.P., t6 = a + (6 - 1)d = a + 5d9th term of the A.P., t9 = a + (9 - 1)d = a + 8dt19 = 3t6{tex}\\Rightarrow{/tex}a + 18d = 3(a + 5d){tex}\\Rightarrow{/tex}a + 18d = 3a\xa0+ 15d{tex}\\Rightarrow{/tex}18d - 15d = 3a - a{tex}\\Rightarrow{/tex}3d = 2a{tex}\\therefore \\mathrm { a } = \\frac { 3 \\mathrm { d } } { 2 }{/tex}t9 = 19{tex}\\Rightarrow \\frac { 3 \\mathrm { d } } { 2 } + 8 \\mathrm { d } = 19{/tex}{tex}\\Rightarrow \\frac { 3 d + 16 d } { 2 } = 19{/tex}{tex}\\Rightarrow \\frac { 19 \\mathrm { d } } { 2 } = 19{/tex}{tex}\\Rightarrow{/tex}\xa0d = 2{tex}\\Rightarrow{/tex}a = 3t2\xa0= 3 + (2 - 1)2 = 5t3\xa0= 3 + (3 - 1)2 = 7The series will be 3, 5, 7...... | |
| 16547. |
cosec A √1- cos² A = 1 |
| Answer» | |
| 16548. |
1/7x+ 1/6y=31/2x-1/3y=5 |
| Answer» {tex}\\frac{1}{7x}{/tex}\xa0+\xa0{tex}\\frac{1}{6y}{/tex}\xa0= 3 .......(i)and\xa0{tex}\\frac{1}{2x}{/tex}\xa0-\xa0{tex}\\frac{1}{3y}{/tex}\xa0= 5 ...........(ii)Multiplying equation (ii) by\xa0{tex}\\frac{1}{2}{/tex}, we get\xa0{tex}\\frac{1}{4x}{/tex}\xa0-\xa0{tex}\\frac{1}{6y}{/tex}\xa0=\xa0{tex}\\frac{5}{2}{/tex} ..........(iii)Adding eq. (i) and (iii), we get{tex}\\frac{1}{4x}{/tex}\xa0+\xa0{tex}\\frac{1}{7x}{/tex}\xa0=\xa0{tex}\\frac{5}{2}{/tex}\xa0+ 3{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{7 + 4}{28x}{/tex}\xa0=\xa0{tex}\\frac{11}{2}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{11}{28x}{/tex}\xa0=\xa0{tex}\\frac{11}{2}{/tex}{tex}\\Rightarrow{/tex}\xa028x = 2\xa0{tex}\\Rightarrow{/tex}\xa0x =\xa0{tex}\\frac{1}{14}{/tex}Putting the value of x in eq.(i), we get{tex}\\frac{1}{7(\\frac1{14})}{/tex}\xa0+\xa0{tex}\\frac{1}{6y}{/tex}\xa0= 3y =\xa0{tex}\\frac{1}{6}{/tex}Hence x = {tex}\\frac{1}{14}{/tex} and y = {tex}\\frac{1}{6}{/tex} is the solution of given system of equations. | |
| 16549. |
Write all trigonometric ratios in terms of sin theta |
| Answer» | |
| 16550. |
Maths exercise 8.4 |
| Answer» | |