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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.
| 16801. |
how to construct an external tangent of a circle with radius 5cm |
| Answer» | |
| 16802. |
What must be subtracted from x3–6x2–15x+80 so that the result is exactly divisible by x2+x–12. |
| Answer» Divide the first expression by second expression . | |
| 16803. |
State the pass marks if 75%of the candidates pass |
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| 16804. |
30 girls do a work in 15 days how many girls will so this work in 9 days |
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| 16805. |
ax-by=a-bbx-ay=a+b |
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Answer» ax + by = a - b multiply by abx - ay = a + b multiply by b{tex}\\Rightarrow{/tex} x = 1{tex}\\therefore{/tex} a + by = a - bby = -by = -1{tex}\\therefore{/tex} x = 1y = -1 Thk तक |
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| 16806. |
Solve for x x + 1 upon x minus 1 plus x minus 2 upon X + 2 equal to 4 minus 2 X + 3 upon x minus 3 |
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| 16807. |
2+2 what answer |
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Answer» 4 4 4... Yup yaar ye kesa question hai????? Yeh kaisa question hai .....??????? 4.. |
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| 16808. |
Who invented pie |
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Answer» A scientist Rahul Gandhi Aryabhatta William Jones |
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| 16809. |
If P is prime then find HCF and LCM of P and (p+1) |
| Answer» HCF =1LCM = product of p and (p+1) | |
| 16810. |
For what value of K is 3 a solution of x 2 +11x+K pl find the value of K |
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Answer» Let f(x)= x2 + 11x + kIf -3 is zero of f(x) then f(-3)=0(-3)2 + 11{tex}\\times{/tex}(-3) + k = 09 - 33 + k = 0- 24 + k = 0k = 24 Putting x = 39 + 33 + k = 0k = -42 |
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| 16811. |
l CM of 313 |
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| 16812. |
lCM of 312 |
| Answer» 2×2×2×3×13 | |
| 16813. |
14માં પાઠ ના કૌશલ્ય ના દાખલા |
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Answer» ???? ?...?? Gyi? Dekh lo sad ke baad happy vala emoji bhi hai na...bs ab 3 days aur phir na mai yaha aungi aur na koi hurt hoga Sad nahi hu mai.."virtual frnds ke liye kaisa sad hona"..yeh mujhe kisi ne sikhaya hai aur ab mai yahi follow karungi bs Yapp Why r u so sad K. Bestie? Diya ???? What\'s the ques.?????.?can u plz write this in English or in hindi!!?? |
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| 16814. |
30 question of chapter 1 math Board exam class 10th |
| Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 16815. |
30 questions of math Board exam class 10th |
| Answer» Cant provide you here...u could refer google | |
| 16816. |
Why x×y=xy Y×z=yz Q×R=qrBut,why 0×4=04 |
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| 16817. |
Find the value of \'K\' which have infinitly many solutions. 1. 2x+3y=k (k-1)x+(k+2)y=3k |
| Answer» {tex}2x + 3y = 7{/tex}{tex}(k - 1) x + (k + 2)y = 3k{/tex}These are of the form{tex}a_1x + b_1y + c_1\xa0= 0 ,\\ a_2x + b_2y + c_2\xa0= 0{/tex}where,{tex}a_1= 2\xa0,\\ b_1= 3,\\ c_1\xa0= -7,{/tex}{tex}a_2=k - 1\xa0\\ ,b_2= k + 2 ,\\ c_2\xa0= -3k {/tex} for infinitely many solutions, we must have{tex}\\frac { a _ { 1 } } { a _ { 2 } } = \\frac { b _ { 1 } } { b _ { 2 } } = \\frac { c _ { 1 } } { c _ { 2 } }{/tex}This hold only when{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 } = \\frac { - 7 } { - 3 k }{/tex}{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 } = \\frac { 7 } { 3 k }{/tex}Now the following cases arises:Case I:{tex}\\frac { 2 } { k - 1 } = \\frac { 3 } { k + 2 }{/tex}{tex}\\Rightarrow{/tex}2(k + 2) = 3(k -1){tex}\\Rightarrow{/tex}2k + 4= 3k - 3{tex}\\Rightarrow{/tex}\xa0k = 7Case II:{tex}\\frac { 3 } { k + 2 } = \\frac { 7 } { 3 k }{/tex}{tex}\\Rightarrow{/tex}7(k + 2) = 9k{tex}\\Rightarrow{/tex}7k + 14= 9k{tex}\\Rightarrow{/tex}\xa0k = 7Case III:{tex}\\frac { 2 } { k - 1 } = \\frac { 7 } { 3 k }{/tex}{tex}\\Rightarrow{/tex}\xa07k - 7 = 6k{tex}\\Rightarrow{/tex}\xa0k = 7For k = 7, there are infinitely many solutions of the given system\xa0of equations. | |
| 16818. |
Express the number 0.3178 in p/q form |
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Answer» But I forgoted to put bar on 3178 3178÷10000 3178/10000 |
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| 16819. |
What about R.D sharam and worksheet question and answer |
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Answer» What you ask from rd sharma i can help you What you want ask? Hii What do u want exactly? |
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| 16820. |
2×3 |
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Answer» 2× 3= 6 What a stupid question. It is a basic maths. Class I students will do better than you . Shame on you. You don\'t deserve to read in class X now . You deserve to read in class I 6 Answer is 6 6 6 Your answer is 6...Hope it is helpful. Your answer is 6. |
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| 16821. |
If m=n and n=m then find the mn term |
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| 16822. |
CosA-sin A +1/cosA + sinA-1=cosecA +cotA |
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| 16823. |
Previous years question of lesson 4 |
| Answer» I v n gn m gn gnn ghh diy d whoop | |
| 16824. |
How to know whether the given equation is quadratic or not |
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Answer» Thank you guys Eguation power has 2 If a equation is of the form ax^2+bx+c=0 and a is not 0 ,a,b,c are real numbers then the equation is a quadratic equation. If the eqation has highest power 2 Quadratic equation is the equation which have degree of 2 &In the form of 2 ax. +bx + c =0 If the given equation has the highest degree 2 then it is a quadratic equation. If the highest power of variable ( degree ) in an eqiation is 2 then it is a quadratic equation. |
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| 16825. |
tan A ÷sec A-1+tan A ÷sec A+1=2 cosec A.prove it |
| Answer» LHS =\xa0{tex}\\frac{\\tan A}{\\sec A-1}+\\frac{\\tan A}{\\sec A+1}{/tex}{tex}=\\frac{\\tan A(\\sec A+1)+\\tan A(\\sec A-1)}{(\\sec A-1)(\\sec A+1)}{/tex}{tex}=\\frac{\\tan A \\cdot \\sec A+\\tan A+\\tan A \\sec A-\\tan A}{\\sec ^{2} A-1}{/tex}{tex}=\\frac{2 \\tan A \\sec A}{\\tan ^{2} A}{/tex}\xa0[{tex}\\because{/tex}\xa0{tex}(sec^2\\theta - 1) = tan^2\\theta\xa0{/tex}]{tex}=\\frac{2 \\sec A}{\\tan A}{/tex}{tex}=\\frac{2 \\frac{1}{\\cos A}}{\\frac{\\sin A}{\\cos A}}{/tex}{tex}=2 \\times \\frac{1}{\\cos A} \\times \\frac{\\cos A}{\\sin A}{/tex}{tex}=\\frac{2}{\\sin \\mathrm{A}}{/tex}{tex}= 2 cosec\\ A{/tex} = RHS | |
| 16826. |
Solve by cross multiplication X/a +y/b =2 A2x + b2y = a3 +b3 |
| Answer» Xxxxx | |
| 16827. |
Solve the eq x_1/×+2 + ×_3/×_4 = 10/3 |
| Answer» What do u mean by _ symbol?????? | |
| 16828. |
Prove that route 5 is irrational |
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Answer» Hence, ROOT 5 is irrational ?????? Let√ 2 be a rational numberTherefore,√2 =a/bWhere a & b are Co-prime integers &b is not = 0Therefore,a/b=√2a= √2bSquaring both sides (a)2=(√2b)2a2= 2b2......(1)This implies 2 divides a2So,2 ➗ aSo let a = 2 c for some integerPut this in (1...)(2c)2 = 2b square4c square= 2b square2c square= b squareThis implies, 2 divides b squareSo,2 divides bTherefor, a & have 2 as commen factor but this contradicts that a & b no common factor then b. Therefore our aaaumption is wrongHence √2 is irrationalThis is the method of doing question you just have to change it from √ 2 to √5 and all over method is same.. Let us assume that √5 is a rational number.we know that the rational numbers are in the form of p/q form where p,q are integers.so, √5 = p/q p = √5qwe know that \'p\' is a rational number. so √5 q must be rational since it equals to pbut it doesn\'t occurs with √5 since its not an integertherefore, p = √5qthis contradicts the fact that √5 is an irrational numberhence our assumption is wrong and √5 is an irrational number.\xa0 This app is for aking homework related questions. And he is doing ✔ This app is for chatting olny Ask to your teacher don\'t ask this type of questions in this app |
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| 16829. |
Find the centre of a circle passsing through a point (6,_6),(3,-7),(3,3) |
| Answer» Let\xa0A → (6, –6), B\xa0→ (3, –7) and C\xa0→ (3, 3).Let the centre of the circle be I(x, y)Then, IA = IB = IC [By definition of a circle]{tex}\\Rightarrow{/tex} IA2 = IB2 = IC2{tex}\\Rightarrow{/tex} (x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2Taking first two, we get(x - 6)2 + (y + 6)2 = (x - 3)2 + (y + 7)2{tex}\\Rightarrow{/tex} x2 - 12x + 36 + y2 + 12y + 36 = x2 - 6x + 9 + y2 + 14y + 49{tex}\\Rightarrow{/tex} 6x + 2y = 14{tex}\\Rightarrow{/tex} 3x + y = 7 ......(1) ....[Dividing throughout by 2]Taking last two, we get(x - 3)2 + (y + 7)2 = (x - 3)2 + (y - 3)2{tex}\\Rightarrow{/tex} (y + 7)2 = (y - 3)2{tex}\\Rightarrow{/tex} (y + 7) = {tex}\\pm{/tex}(y-3)taking +e sign, we gety + 7 = y - 3{tex}\\Rightarrow{/tex} 7 = -3which is impossibleTaking -ve sign, we gety + 7 = -(y - 3){tex}\\Rightarrow{/tex} y + 7 = -y + 3{tex}\\Rightarrow{/tex} 2y = -4{tex}\\Rightarrow y = \\frac{{ - 4}}{2} = - 2{/tex}Putting y = -2 in equation (1), we get{tex}\\Rightarrow{/tex} 3x - 2 = 7{tex}\\Rightarrow{/tex} 3x = 9{tex}\\Rightarrow{/tex} x = 3Thus, I {tex}\\rightarrow{/tex} (3, -2)Hence, the centre of the circle is (3, -2). | |
| 16830. |
A+B+C=90 then summation cos(B+C)/cosB. cosC |
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| 16831. |
Solvethe system of equationx/2+y=0.8. 7/x+y/2=10. |
| Answer» Given equations are{tex}\\frac{x}{2} + y = 0.8{/tex}{tex}\\frac{x + 2y}{2} = 0.8{/tex}.x + 2y = 1.6 ........ (i)and{tex}\\frac{7}{{x + \\frac{y}{2}}} = 10{/tex}\xa0{tex}\\frac{{7 \\times 2}}{{2x + y}} = 10{/tex}{tex}\\frac{{7}}{{2x + y}} = 5{/tex} 7 = 10x + 5y10x + 5y = 7 .......... (ii)Multiply first equation by 1010x + 20y = 16............ (iii)Subtracting the equations (ii) from (iii) , we get15y = 9{tex}y = \\frac{9}{{15}} = \\frac{3}{5}{/tex}Put value of y in (i){tex}x = 1.6 - 2\\left( {\\frac{3}{5}} \\right) = 1.6 - \\frac{6}{5} = \\frac{2}{5}{/tex}Solution is\xa0{tex}\\left( {\\frac{2}{5},\\;\\frac{3}{5}} \\right){/tex} | |
| 16832. |
what ia use of trigonometry |
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Answer» This is use fkr measuring the 3 triangle for measuring the three angle |
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| 16833. |
Are there different subject codes for maths basic and standard in cbse class 10? |
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Answer» Don\'t know ?? no ... |
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| 16834. |
Solve 2x +3y=11 amd 2x-4y = -24 and hence find tje value of m for which y=mx+3 |
| Answer» m=-1 | |
| 16835. |
Solve system of equation by cross multiplication method. mx - ny =m square +n square, x +y = 2m |
| Answer» Given equations aremx - ny = m2 + n2 .................(i)x + y = 2m ....................(ii)Herea1 = m, b1 = -n, c1 = -(m2 + n2)a2 = 1, b2 = 1, c2 = -(2m)By cross multiplication method{tex}\\frac{x}{{{{( 2mn + m^2 + n^2)} }}} = \\frac{{ - y}}{{ - {2m^2} + m^2 + n^2 }} = \\frac{1}{{m + n}}{/tex}{tex}\\frac{x}{{{{(m + n)}^2}}} = \\frac{{ - y}}{{ - {m^2} + {n^2}}} = \\frac{1}{{m + n}}{/tex}Now, {tex}\\frac{x}{{{{(m + n)}^2}}} = \\frac{1}{{m + n}} {/tex}{tex}⇒ x = m + n{/tex}And, {tex}\\frac{{ - y}}{{ - {m^2} + {n^2}}} = \\frac{1}{{m + n}}{/tex}{tex} ⇒ y = m - n{/tex}The solutions of the given pair of equations are x = m + n and y = m - n. | |
| 16836. |
Jd8d |
| Answer» ......?? | |
| 16837. |
If Sn denotes the sum of first n terms of an AP,prove thatS12=3 (S8_S4) |
| Answer» Let a be the first term and d be the common difference of the given AP. Then,Sn= {tex}\\frac{n}{2}{/tex}{tex} \\cdot {/tex}[2a+(n-l)d],{tex}\\therefore{/tex}\xa0{tex}3(S_8-S_4) = 3{/tex}[{tex}\\frac{8}{2}{/tex}{tex}(2a+7d)-{/tex}{tex}\\frac{4}{2}{/tex}{tex}(2a+3d)]{/tex}= {tex}3[4(2a+7d)- 2(2a+3d)] = 6(2a+11d){/tex}{tex}= \\frac { 12 } { 2 } \\cdot ( 2 a + 11 d ) = S _ { 12 }{/tex}.Hence, S12= 3(S8-S4). | |
| 16838. |
321/4 |
| Answer» 80.25 | |
| 16839. |
Exercise 2 ,2 4s-4s+1 |
| Answer» 4s-(2+2)s+14s-2s-2s+1 2(2s-1)-1(2s-1)(1)(2s-1) | |
| 16840. |
if HCF (12,15) =3, find LCM (12,15). |
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Answer» Lcm= 60 LCM×HCF= a×b,LCM×3=12×15,LCM=180/3=60. Therefore, LCM= 60. |
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| 16841. |
Ten. 90 decre |
| Answer» tan 90°{tex} = \\infty{/tex} | |
| 16842. |
Find the 15th term and 22nd term of the AP x+b,x+3b,x+5b |
| Answer» The Given Series : x+b, x+3b, x + 5b,...This is an A.P.∴ First Term, a = x+bDifference, d = x+3b - (x+b) = x+3b - x-b= 2bNow, The General Term ,= a+(n-1)d= x+b+(n-1)2b= x+b+(2n-2)b= x+b(1+2n-2)= x+b(2n-1)a = x+b | |
| 16843. |
In an AP, given l=28, S=144 and there are total 9 terms find a |
| Answer» l = 28S = 144n = 9We have to find a,The value of a = 4 | |
| 16844. |
Divide 22222$ |
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| 16845. |
Proof that root 5 is irrational |
| Answer» Let us assume that √5 is a rational number.we know that the rational numbers are in the form of p/q form where p,q are integers.so, √5 = p/q p = √5qwe know that \'p\' is a rational number. so √5 q must be rational since it equals to pbut it doesnt occurs with √5 since its not an integertherefore, p =/= √5qthis contradicts the fact that √5 is an irrational numberhence our assumption is wrong and √5 is an irrational number.\xa0 | |
| 16846. |
I want 2019-2020 qestion paper |
| Answer» Check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 16847. |
I want the solutions of RS aggarwal |
| Answer» Which one??..provide the math please.. | |
| 16848. |
Hello guys any one is online????i am kaya |
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Answer» How are you ?? Hello... Hello Hi |
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| 16849. |
show that one and only one out of n,n+2,n+4 is divisible by 3 ,where n is any positive integer |
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Answer» We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0 | |
| 16850. |
show that 1/3+5√2 is an irrational |
| Answer» | |