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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.
| 41951. |
If sinA+cosA=2 then find tanA+cotA= |
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Answer» Your answer is 2/3 Simplify tanA + cotA in terms of sinA and cosA U will get 1/sinA . 1/cosASinA + cosA = 2Squaring both sides givesSin^2A + cos^2A + 2sinA.cosA = 41 + 2 sin A.cosA = 4SinA. CosA = 3/21/sinA. 1/cosA = 2/3TanA + cotA = 2/3 |
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| 41952. |
Explain BPT theorem. |
| Answer» Its given in ncert, plz check there, as its too long to type here, hope yu can understand ? | |
| 41953. |
P(x)=(k+4)x2+13x+3k find value of k |
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Answer» You need to have a solution then you may find value of k Please give answer with complete solution 1 |
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| 41954. |
Show that square of an odd digit positive integer is of the form 8m+1,for some integer |
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Answer» Let a be the odd positive intezer and b=8.By Euclid division lemma where r=0,1,2,3,4,5,6,7.When r=0a=8q+0When r=1a=8q+1 (odd)When r=2a=8q+2When r=3a=8q+3 (odd)When r=4a=8q+4When r=5a=8q+5 (odd)When r=6a=8q+6When r=7a=8q+7 (odd) According to Euclid division lemma , a = bq + r where 0 ≤ r < bHere we assume b = 8 and r ∈ [1, 7 ] means r = 1, 2, 3, .....7Then, a = 8q + rCase 1 :- when r = 1 , a = 8q + 1squaring both sides,a² = (8q + 1)² = 64²q² + 16q + 1 = 8(8q² + 2q) + 1= 8m + 1 , where m = 8q² + 2qcase 2 :- when r = 2 , a = 8q + 2squaring both sides,a² = (8q + 2)² = 64q² + 32q + 4 ≠ 8m +1 [ means when r is an even number it is not in the form of 8m + 1 ]Case 3 :- when r = 3 , a = 8q + 3squaring both sides,a² = (8q + 3)² = 64q² + 48q + 9 = 8(8q² + 6q + 1) + 1= 8m + 1 , where m = 8q² + 6q + 1You can see that at every odd values of r square of a is in the form of 8m +1But at every even Values of r square of a isn\'t in the form of 8m +1 .Also we know, a = 8q +1 , 8q +3 , 8q + 5 , 9q +7 are not divisible by 2 means these all numbers are odd numbersHence , it is clear that square of an odd positive is in form of 8m +1.\xa0 |
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| 41955. |
Find the 22th term of an ap1,2....4 |
| Answer» 22 | |
| 41956. |
Difference between altitude and median |
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Answer» Median\tA line segment passing through a vertex to the midpoint of an opposite side is called a median.\tThe point where the 3 medians meet is called the centroid of the triangle.\tEach median of a triangle divides the triangle into two smaller triangles which have equal area.Altitude\tAn altitude passing through a vertex and is at right-angle at the opposite side.\tThe point where the 3 altitudes meet is called the ortho-center of the triangle.\tThe altitude of a triangle may lie inside or outside the triangle. Median divides the triangle into two equal parts whereas altitude makes 90 degrees Altitude is always perpendicular or inclined at right angle but median is a line which divide aline in ratio 1:1 |
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| 41957. |
Find hcf and lcm of 404 and 96 and verify that hcf *lcm=product of two given numbers. |
| Answer» Hcf is 4. | |
| 41958. |
If a point A (0,2) is equidistant from the point B (3,p) and C (p,5)then find the value of p |
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Answer» AB=ACAB^2=AC^2(0-3)^2+(2-p)^2=(o-p)^2+(2-5)^29+4+p^2-4p=p^2+9-4p=-4p=1 P=1 |
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| 41959. |
Find the value of p in equation px×x+2x-3 |
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Answer» Pl check your question is not complete.If roots are equal thenb^2=4ac2^2=4 x p x -3p=-1/3\xa0\xa0 Px x x + 2x - 3Px sq. + 2x -3You\'ll get a quadratic eq. Now u can use quadratic formula, middle term split and completing the squareOut of three u can use any method for finding the value of xI hope u got ur answer....? |
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| 41960. |
Find the value of p if the point a(0;2)is equidistant from (3;p) and(p;3). |
| Answer» P=3 | |
| 41961. |
If cos 47\' = a ; then find the values of sin43\' and tan 43\' in terms of a |
| Answer» sin43=sin(90-47)=cos47=a{tex}\\begin{array}{l}cot43=\\sqrt{1+}\\cos ec^243\\\\=\\sqrt{1+\\frac1{a^2}}=\\frac1a\\sqrt{1+a^2}\\\\\\tan43=\\frac1{cot43}=\\frac a{\\sqrt{1+a2}}\\end{array}{/tex} | |
| 41962. |
If ABC-AORP, arRE.and BC -15 cm, then find PR. |
| Answer» {tex}\\frac{\\operatorname{arc}(\\triangle \\mathrm{ABC})}{\\operatorname{ar}(\\triangle \\mathrm{QRP})}{/tex}\xa0=\xa0{tex}\\left(\\frac{B C}{R P}\\right)^{2}{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{9}{4}{/tex}\xa0=\xa0{tex}\\left(\\frac{15}{P R}\\right)^{2}{/tex}{tex}\\Rightarrow{/tex} PR = 10 cm | |
| 41963. |
Express An of an AP in terms of Sn and Sn-1 |
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Answer» An = Sn-S( n-1) An= Sn-S(n-1) |
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| 41964. |
Surface area and volume all formula |
| Answer» CUBOID:-(1) VOLUME OF CUBOID = L×B×H(2) CSA OF CUBOID= 2(L+B)×H(3) TSA OF CUBOID= 2(LB+BH+LH)CUBE:-(1) VOLUME OF CUBE = A³(2) CSA OF CUBE= 4A²(3) TSA OF CUBE = 6A²CYLINDER:-(1) VOLUME OF CYLINDER= πR²H(2) CSA OF CYLINDER= 2πRH(3) TSA OF CYLINDER= 2πR(H+R)CONE:-(1) VOLUME OF CONE = 1/3πR2H(2) CSA OF CONE = πRL (3) TSA OF CONE = πRL +πR² = πR(L+R)SPHERE:-(1) VOLUME OF SPHERE = 4/3πR³(2) TSA OF SPHERE = 6A²HEMISPHERE :-(1) VOLUME OF HEMISPHERE = 2/3πR³(2) CSA OF HEMISPHERE = 2A²(3) TSA OF HEMISPHERE = 3A².FRUSTUM OF CONE:-(1) VOLUME = R²+r²+ Rr(2) CSA = πL(R+r)²Read more on Brainly.in - https://brainly.in/question/2418027#readmore | |
| 41965. |
If 1+5+7+10+......+x = 287 find the value of x |
| Answer» Given, 1 + 4 + 7 + 10 +...+ x = 287This series in AP.Now, first term a = 1, common difference d = 4 - 1 = 3Last term l = xLet number of terms = nNow, Sum of the seires Sn = 287=> (n/2) ×{2a + (n - 1)d} = 287=> (n/2) ×{2 + (n - 1)3} = 287=> (n/2) ×{2 + 3n - 3} = 287=> (n/2) ×{3n - 1} = 287=> n ×(3n - 1) = 287 ×2=> 3n2 - n = 574=> 3n2 - n - 574 = 0=> (n - 14) ×(3n + 41) = 0=> n = 14, -41/3Since n can not be negative,So, n = 14i.e. there are 14 terms in the series and x is the 14th term.So, x = a14 = a + (14 - 1)d=> x = a + 13d=> x = 1 + 13 ×3=> x = 1 + 39=> x = 40So, the value of x is 40 | |
| 41966. |
Solve the equation.... 1+4+7+10+.......x=287.... Question from ap |
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Answer» x=14 Honey right X is 14th term and the value of x is 40 No its 14 40 |
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| 41967. |
2018 question paper maths |
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Answer» In CBSE guide,last year question paper in maths subject with solution On Google |
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| 41968. |
Can I get a sample in which the name of chapter in particular question is mentioned?? |
| Answer» | |
| 41969. |
If ratio of sum of 1st n terms of 2 AP is 7n+1:4n+27. Find ratio of 9th term |
| Answer» Let the sum of first n terms of two AP be S1 and S2.Then , S1/ S2 = | |
| 41970. |
The nth termof an AP is given by-4n+15.find the sum of first 20 terms |
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Answer» A1=-4(1)+15=11, A2=-4(2)+15=7 , D=7-11=-4. S20=20/2(22+19(-4))=10(22-76)=10(-54)=-540 95 an=-4n+15a1=-4×1+15 =11a2=-4×2+15 =7CD=7-11 =-4a20=a+19d =11+19×-4 =11-76 =-65 May be..... 18 |
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| 41971. |
Circumcentre of circle |
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Answer» One of several centers the triangle can have, the\xa0circumcenter\xa0is the point where the perpendicular bisectors of\xa0a triangle\xa0intersect. The\xa0circumcenter\xa0is also the center of the triangle\'s circumcircle - the\xa0circle\xa0that passes through all three of the triangle\'s vertices. When we draw three perpendicular bisector from three sides of triangle and where they meet inside the triangle inside the triangle is called circumcenter. 2πr |
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| 41972. |
(1-b)²×a×(1+b)²=-7/2 |
| Answer» ?? | |
| 41973. |
Cosec13°/sec77°-cot20°/tan70° |
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Answer» Its 0 0 |
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| 41974. |
Sec^2Q cosec^2Q -2-cot^2Q =tan^2Q |
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Answer» How to give yu solution here.......? Yes |
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| 41975. |
Find the circumference of a circle is 22cm.the area of quadrant in cm |
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Answer» Answer is 38.5.... 9.625 Circimference of a circle=22 cm 2πr=22 2×π×r=22 r=22×7÷2×7 r=3.5 cm Area of quadrant= π×r×r÷4 =π×3.5×3.5÷4 = 154 cm sq. 38.5 cm2 |
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| 41976. |
1 is prime no. or composite no. |
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Answer» It is neither prime nor composite.. 1 is neither prime nor composite number |
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| 41977. |
Triangle ABC~trianglePQR AB/PQ=1/3 find ar(ABC)/ar(PQR) |
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Answer» Answer is 1/9 1/9 square of corresponding sides |
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| 41978. |
Xsqure-3x-m(m+3)=0 m is constant |
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| 41979. |
Xsqure+x-p(p+1)=0 p is constsnt |
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| 41980. |
In a isosceles triangle ABC where AB = AC and BD perpendicular AC To prove BD^2 - CD^2 = 2CD.AD |
| Answer» | |
| 41981. |
Tow chords AB&CD intersect each other at point p . Prove that triangleAPC similar to triangle DPB |
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Answer» Your questin was right Given: AB&CD are chords.PT:∆APC~∆DPBconstruction: Join AC&BD such that AC||BD.Prove: In∆APC &DPB : |
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| 41982. |
tan ^2 + cot ^2 = 2 then tan + cot = |
| Answer» | |
| 41983. |
3+2root3 |
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Answer» 6.464 Abcose irrational no. Bro 3+2√3is a irrational no. I think that\'s what you want.... |
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| 41984. |
Anyone knows about alternative segment theorm |
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Answer» The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment. In any circle, the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment, i.e. the angle subtended by the chord in the opposite side of the previous angle. No please also tell me |
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| 41985. |
What are rational |
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Answer» A rational number is a number which vam ne written in the form of p/q where q is equal not 0. A rational number (Q) is any number which can be written as: a/b , where a and b are integers and b≠0. The rational number is the number which can be written in the form of p/q where q is not equal to zero |
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| 41986. |
Synch [email\xa0protected]@7# |
| Answer» What is the question | |
| 41987. |
A boat is |
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Answer» A boat is ???????????????? ????? Further question What a boat is..?????? |
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| 41988. |
(Sin3A +cos3A )+(sinA cosA )(sinA+cosA )/sinA +cosA |
| Answer» Plz write the whole Question. | |
| 41989. |
Marking scheme of chapter 13 of class 10 |
| Answer» There are mainly 4 or 3 marks questions come in board exam. | |
| 41990. |
How to justifyconstruction of similiar triangles? |
| Answer» There is no need to justify construction. | |
| 41991. |
Find the Roots of the equation x+√x-2=4 |
| Answer» X + √(x - 2) = 4\xa0√(x - 2) = 4 -x\xa0take square Both sides\xa0x - 2 = 16 -8x + x²x² -9x +18 = 0x² -6x -3x +18 =0x(x -6) -3(x -6) =0(x -6)(x -3) =0x = 6 , 3\xa0according to √(x - 2) and (4 -x) ≥0x≥2 and x ≤4\xa0so, x ≠ 6\xa0only x = 3 | |
| 41992. |
If x=2/3 and x=-3 are the roots of ax^2+ bx+ c find a and b |
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| 41993. |
volume of cone = ...........volume of cylinder |
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Answer» Mr Guar behave like class x student otherwise it looks likeyour mind is still like a primary class student.The person like me will never answer your question 1/3πr^²h 1/3 of cyclinder πr2h 1/3 |
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| 41994. |
Formula of cylinder |
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Answer» Volume of cylinder is 22/7.r.r.h &c.s.a of cylinder is 2.22/7r.h &t.s.a of cylinder is 2.22/7.r.( r + h) C.S.A/L.S.A=2ΠrhVol:Πr²h See below |
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| 41995. |
What are the possible values of remainder r, when a positive integer a is divided by 3 ? |
| Answer» 0,1 and 2..... Since b=3 0<=r<3 | |
| 41996. |
2 X square + 3 X + 4 is quadratic equation |
| Answer» Yes it is a quadratic equation | |
| 41997. |
How images of question is sent in this app |
| Answer» We can\'t send images of question in this app. | |
| 41998. |
Sir ham is app me sum ki photo kaise bhej sakte ha |
| Answer» | |
| 41999. |
If a√b + b√a = 183 and a√a + b√b = 182 then find the value of 9/5(a+ b) |
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| 42000. |
6+9-2+6+7-10=?? |
| Answer» | |