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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.
| 17051. |
Q .If a dice is tossed 3 times then find the probability getting at least 3 on a dice |
| Answer» I assume you mean “IF a die is tossed three times, what is the probability of no fives?” Die #1 can do so with probability 5 / 6 (meaning, no fives). Ditto for #2, #3. Hence, the required probability is (5/6)*(5/6)*(5/6) = 125/216, which is almost 58%. | |
| 17052. |
If √3tantheta =1then evaluate (cos^2theta - Sin^2theta). |
| Answer» 1/2 | |
| 17053. |
If root3 tan theta is equal to 3sin theta then prove that cos square theta-sin square theta |
| Answer» | |
| 17054. |
Math s |
| Answer» What maths | |
| 17055. |
(Cot-cosec) ^2=1-cos/1+cos |
| Answer» LHS,(cot-cosec)²=(cos/sin-1/sin)²=(cos-1/sin)²=cos²+1-2cos/sin²=2-2cos-1+cos²/1-cos²=2(1-cos)-1(1-cos²)/1-cos²=2(1-cos)-1((1-cos)(1+cos))/(1-cos)(1+cos)=(1-cos)(2-1-cos)/(1-cos)(1+cos)Hence,(1-cos)/(1+cos) prove. | |
| 17056. |
Tan² theta -sin2 theta |
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Answer» Sin square theta/ cos square theta- sin square theta. (1/cos square theta - 1). Sin square theta. (Sec square theta - 1) . Sin square theta. As 1+tan square theta= sec square theta. Sotan square theta. Sin square theta. Hence proved. = |
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| 17057. |
Case study based questions |
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Answer» Give me answer the value of ( 1+ cot²2x)sin²2x is if x= 30⁰ मान ज्ञात करो |
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| 17058. |
Draw two tangent to a circle of radius 3.5 cm from a point P at a distance of 6.2 cm from its centre |
| Answer» Steps of Construction\xa0Step 1: Draw a circle with O as center and radius 3.5 cm.Step 2: Mark a point P outside the circle such that OP = 6.2 cm\xa0Step 3: Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.\xa0Step 4: Draw a circle with Q as center and radius PQ (or OQ), to intersect the given circle at the points T and T’.\xa0Step 5: Join PT and PT’.Here, PT and PT’ are the required tangents.Read more on Sarthaks.com - | |
| 17059. |
9cos2o+9sin20=? |
| Answer» 9(cos2o+sin2o)9(1)9 | |
| 17060. |
Find resistance between 2 ohm, 3 ohm, 6ohm, 4ohm, 10ohm and 5ohm |
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Answer» In series or parallel combination. In series or parallel In parallel or serie combination? In series or parallel ? |
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| 17061. |
The value of (sin 30°+ cos 30°) - (sin 60°+ cos 60°) is |
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Answer» Sin 30° - 1/2 Cos 30° - √3/2Sin 60° - √3/2 Cos60° - 1/2(1/2 + √3/2) - (√3/2 + 1/2)(√3 +1)/2 - (√ 3+1)/2 = 0 (sin30+cos30) - (sin60+cos60)\xa0sin30 = 1/2\xa0sin60= √3/2\xa0cos30= √3/2\xa0cos60= 1/2\xa0=> (1/2+√3/2) - (√3/2+1/2)\xa0=> 1+√3/2 - √3+1/2\xa0=> 1+1+√3-√3 /2\xa0=> 2/2\xa0=>1 |
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| 17062. |
Ex.13.1 ques no.1 |
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Answer» Ff. Hidh ghgj the attitude of the following sheets and towels for class 5 class XThe new time to explore your voice mail system expanded from class X social science Unit 2 chapter 1 Length\xa0=22cm=lθ=60∘180∘=πrad1∘=180π\u200brad60∘=180π\u200b×60rad=3π\u200bradArc length\xa0=rθr=θl\u200b=(3π\u200b)22\u200b=(722\u200b)22×3\u200br=3×7r=21cm A piece of wire 22 cm long is bent into the form of an arc of a circle subtending\xa0an angle of 60o\xa0at its centre. Find the radius of the circle. |
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| 17063. |
Case study questions |
| Answer» | |
| 17064. |
Tan²^-sec²^ = |
| Answer» 1 | |
| 17065. |
Prove that 1-cos square theta)cosec square theta =1 |
| Answer» (1-cos square theta)cosec square theta=(Sin square theta)cosec square theta=Sin square theta ×1/sin square theta=1 | |
| 17066. |
ax²+bx+c =0 |
| Answer» Yes this is true..... But what is ur question? | |
| 17067. |
Can anyone give me deleted Chapters of Maths ? Plz it\'s urgent ? |
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Answer» Thanks you CHAPTER TOPICS REMOVED UNIT I-NUMBER SYSTEMS REAL NUMBERS \uf0b7 Euclid’s division lemma UNIT II-ALGEBRA POLYNOMIALS \uf0b7 Statement and simple problems on division algorithm for polynomials with real coefficients. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES \uf0b7 cross multiplication method QUADRATIC EQUATIONS \uf0b7 Situational problems based on equations reducible to quadratic equationsARITHMETIC PROGRESSIONS \uf0b7 Application in solving daily life problems based on sum to n termsUNIT III-COORDINATE GEOMETRY COORDINATE GEOMETRY \uf0b7 Area of a triangle. UNIT IV-GEOMETRY TRIANGLES Proof of the following theorems are deleted \uf0b7 The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. \uf0b7 In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angle opposite to the first side is a right angle.CIRCLES No deletion CONSTRUCTIONS \uf0b7 Construction of a triangle similar to a given triangle. UNIT V- TRIGONOMETRY INTRODUCTION TO TRIGONOMETRY \uf0b7 motivate the ratios whichever are defined at 0o and 90oTRIGONOMETRIC IDENTITIES \uf0b7Trigonometric ratios of complementary angles.HEIGHTS AND DISTANCES No deletion UNIT VI-MENSURATION AREAS RELATED TO CIRCLES \uf0b7 Problems on central angle of 120° SURFACE AREAS AND VOLUMES \uf0b7 Frustum of a cone. UNIT VI-STATISTICS & PROBABILITY STATISTICS \uf0b7 Step deviation Method for finding the mean \uf0b7 Cumulative Frequency graph PROBABILITY No deletion |
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| 17068. |
Find the ratio in which the line segment joining the points (-6,10) and (3,-8) is divided by (-4,6) |
| Answer» Given:−\u200b\\: \\: \\: \\: \\: \\: \\: P \\: {({-4}, \\: {6})} \\: \\: divides \\: \\: the \\: \\: line \\: \\: segment \\: \\: joining \\: \\: the \\: \\: point \\: \\: A \\: {({-6}, \\: {10})} \\: \\: and \\: \\: B \\: {({3}, \\: {-8})}P(−4,6)dividesthelinesegmentjoiningthepointA(−6,10)andB(3,−8)\\underline\\mathfrak{To \\: \\: Find:-}ToFind:−\u200b\\: \\: \\: \\: \\: find \\: \\: the \\: \\: ratio.?findtheratio.?\\underline\\mathfrak{Solutions:-}Solutions:−\u200b\\: \\: \\: \\: \\: \\: \\: {x} \\: \\: \\: = \\: \\: \\: {-4} \\: \\: \\: \\: \\: \\: \\: \\: {y} \\: \\: \\: = \\: \\: \\: {6}x=−4y=6\\: \\: \\: \\: \\: \\: \\: {x_1} \\: \\: \\: = \\: \\: \\: {-6} \\: \\: \\: \\: \\: \\: \\: \\: {y_1} \\: \\: \\: = \\: \\: \\: {10}x1\u200b=−6y1\u200b=10\\: \\: \\: \\: \\: \\: \\: {x_2} \\: \\: = \\: \\: {3} \\: \\: \\: \\: \\: \\: \\: \\: {y_2} \\: \\: = \\: \\: {-8}x2\u200b=3y2\u200b=−8\\: \\: \\: \\: \\: \\fbox{{x} \\: \\: = \\: \\: \\frac{{mx_2} \\: + \\: {nx_1}}{m \\: + \\: n} \\: \\: \\: \\: \\: \\: \\: {y} \\: \\: = \\: \\: \\frac{my_2} \\: + \\: {ny_1}{m \\: + \\: n}}\\: \\: \\: \\: \\: \\therefore {x} \\: \\: = \\: \\: \\frac{{mx_2} \\: + \\: {nx_1}}{m \\: + \\: n}∴x=m+nmx2\u200b+nx1\u200b\u200b\\: \\: \\: \\: \\: \\leadsto {-4} \\: \\: = \\: \\: \\frac{m \\: {3} \\: + \\: n \\: {-6}}{m \\: + \\: n}⇝−4=m+nm3+n−6\u200b\\: \\: \\: \\: \\: \\leadsto {-4m} \\: - \\: {4n} \\: \\: = \\: \\: {3m} \\: + \\: {(-6n)}⇝−4m−4n=3m+(−6n)\\: \\: \\: \\: \\: \\leadsto {-4m} \\: - \\: {4n} \\: \\: = \\: \\: {3m} \\: - \\: {6n}⇝−4m−4n=3m−6n\\: \\: \\: \\: \\: \\leadsto {-4m} \\: - \\: {3m} \\: \\: = \\: \\: {-6n} \\: + \\: {4n}⇝−4m−3m=−6n+4n\\: \\: \\: \\: \\: \\leadsto {-7m} \\: \\: = \\: \\: {-2n}⇝−7m=−2n\\: \\: \\: \\: \\: \\leadsto \\frac{m}{n} \\: \\: = \\: \\: \\frac{-2}{-7}⇝nm\u200b=−7−2\u200b\\: \\: \\: \\: \\: \\leadsto {m} \\: : \\: {n} \\: \\: = \\: \\: {2} \\: : \\: {7}⇝m:n=2:7\\: \\: \\: \\: \\: \\therefore {x} \\: \\: = \\: \\: \\frac{{my_2} \\: + \\: {ny_1}}{m \\: + \\: n}∴x=m+nmy2\u200b+ny1\u200b\u200b\\: \\: \\: \\: \\: \\leadsto {6} \\: \\: = \\: \\: \\frac{m \\: {-8} \\: + \\: n \\: {10}}{m \\: + \\: n}⇝6=m+nm−8+n10\u200b\\: \\: \\: \\: \\: \\leadsto {6m} \\: - \\: {6n} \\: \\: = \\: \\: {-8m} \\: + \\: {10n}⇝6m−6n=−8m+10n\\: \\: \\: \\: \\: \\leadsto {6m} \\: - \\: {8m} \\: \\: = \\: \\: {10n} \\: - \\: {6n}⇝6m−8m=10n−6n\\: \\: \\: \\: \\: \\leadsto {14m} \\: \\: = \\: \\: {4n}⇝14m=4n\\: \\: \\: \\: \\: \\leadsto \\frac{m}{n} \\: \\: = \\: \\: \\frac{2}{7}⇝nm\u200b=72\u200b\\: \\: \\: \\: \\: \\leadsto {m} \\: : \\: {n} \\: \\: = \\: \\: {2} \\: : \\: {7}⇝m:n=2:7 | |
| 17069. |
In ∆ABC , right angled at b , ab = 24 cm and BC = 7 cm . determine Sin C |
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Answer» 7/24 is the answer \xa0In\xa0Δ\xa0ABC, B is at right angle.Given, AB=24cmBC=7cmusing Pythagoras theorem\xa0AB2+BC2=AC2⇒(24)2+(7)2=AC2⇒AC=(242)+(7)2\u200b=576+49\u200b=625\u200b⇒AC=25CMsin A=BC\u200b/AC= 7\u200b/25 |
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| 17070. |
marking scheme of mathematics CBSE paper 2021 |
| Answer» CBSE Class 10 Sample Papers 2021 - The Central Board of Secondary Education (CBSE) has released new sample papers with the marking scheme for the upcoming Board Exam 2021. All these sample papers are prepared according to the reduced CBSE syllabus. With the release of the CBSE sample papers, preparations for the board exam will take a kick start as students can now actually prepare according to the pattern followed in the sample papers.\xa0\tCBSE Sample Question PaperCBSE Marking SchemeCBSE Class 10 Basics Maths Sample Paper 2021CBSE Class 10 Basics Maths Marking Scheme 2021CBSE Class 10 Standard Maths Sample Paper 2021CBSE Class 10 Standard Maths Marking Scheme 2021\t | |
| 17071. |
Three vertices of a parallelogram taken in order are (0,3) (0,6) and (2,9). Find the fourth vertex. |
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Answer» 2,6 Let A(- 1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order.Since, the diagonals of a parallelogram bisect each other.So, coordinate of the mid point of AC = coordinate of mid point of BD⇒ [(-1 + 2)/2, (0 + 2)/2] = [(3 + x)/2, (y + 1)/2]⇒ (1/2, 1) = [(3 + x)/2, (y + 1)/2](3 + x)/2 = ½ ⇒ x = - 2Also (y + 1)/2 = 1 ⇒ y + 1 = 2⇒ y = 1The fourth vertex of parallelogram = (- 2, 1). (2,6) |
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| 17072. |
IF x-1?=3, then ? = _______ |
| Answer» | |
| 17073. |
All Formula of real numbers |
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Answer» HCF of two positive integers can be find using the Euclid’s Division Lemma algorithmWe know that for any two integers a. b. we can write following expressiona=bq + r ,\xa00≤r |
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| 17074. |
solve 16+ under root 3 upon 4= 4-1upon2 cos theta |
| Answer» | |
| 17075. |
The sum of the third and fifth terms of an AP is 42 and their product is 432. Find the AP |
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Answer» My friends are going to the market My friends are going to the market. |
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| 17076. |
Find the 7th term from the end of the AP 7, 10, 13............. 184 |
| Answer» Given:- a=7, d=10-7=3, n=7We know that an=a+(n-1)×da7=7+(7-1)×3a7=7+6×3a7=7+18a7=25 Hence, the 7th term of an AP is 25 Ans | |
| 17077. |
If CosA= 12/5 find the value of CosecA-CotA/CosecA+CotA. |
| Answer» 132 +2 | |
| 17078. |
If k+1,3,4k+2 be any three consecutive terms of an AP find the value of k? |
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Answer» K=3/5 hoga 1st term t1 = (k + 1).2nd term t2 = 3k.3rd term t3 = (4k + 2).Given that t1,t2,t3 are in AP.We know that when they are in AP, their common difference will be:t2 - t1 = t3 - t23k - (k + 1) = (4k + 2) - 3k3k - k - 1 = 4k + 2 - 3k2k - 1 = k + 22k = k + 2 + 12k = k + 32k - k = 3k = 3.Therefore the value of k = 3. |
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| 17079. |
Find the distance of the point (-3,4)from the x -axis |
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Answer» 4 cm 5cm Answer : 4 cm |
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| 17080. |
Who is a president of America |
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Answer» Joec Biden is the next America pM Now, Donald TrumpNext, Joe Biden Joe Biden is the current PM of America Joe Biden Joe Biden |
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| 17081. |
A card is drawn from a well shuffled deck of 52 cards.The probability of a seven of spade is |
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Answer» 1/52 No of 7 of spade cards = 1 Total no of cards = 52Therefore probability of 7 of spade is 1/52 1/52 |
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| 17082. |
write 23.426 bar in the form of p by q |
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Answer» Let\'s take 23.426 (repeating on 426) as x.Three digits are1000x - x = 23426.426... - 23.426…999x = 23403x = 23403 / 999 in p/q form Let\'s take 23.426 (repeating on 426) as x.Three digits are repeating by.Multiply by 1000 with x.1000x = 23426.426...x = 23.426Subtracting them,1000x - x = 23426.426... - 23.426…999x = 23403x = 23403 / 999 in p/q form. |
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| 17083. |
Are the indentitiesSin²+Cos²=1 andSin+Cos=1. Are same.Pls tell me |
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Answer» No It is not equal No it\'s not equal No, sin²A + cos²A = 1 but sinA + cosA≠1 |
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| 17084. |
Given that ap% A1,A2, A3,.... For which A1 =1,d=4, find a8. |
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Answer» A1 + 7d = 1 + 7×4 = 29 A8 = 29 29 |
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| 17085. |
Express the logarithms of the following into sum of the logarithms |
| Answer» Idk | |
| 17086. |
The pair of equation y :3 and y :8 has one solution or unique solution or |
| Answer» Unique solution | |
| 17087. |
2+2=5 how???? |
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Answer» By mistake Galti se It is not possible 20 - 20 = 25 - 25 (0=0)(4*5) - (4*5) = (5*5) - (5*5)Taking 4 common from LHSAnd 5 from RHS4(5-5) = 5(5-5)Cancelling (5-5) from both sides4= 54 can be written as 2+2.: 2+2 = 5You might have seen this on internet ,But we cannot prove that 4 = 5.In the third step we can observe that 5-5 is 0 .: 4(0) = 5(0)0 = 0Any no. Multiplied by 0 gives zeroOr we can write it as 4/0 = 5/0Which is infinity. |
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| 17088. |
5=a+2d |
| Answer» So...wht.... | |
| 17089. |
In a triangle ABC & triangle DEF, |
| Answer» Write the full questions please | |
| 17090. |
IT tan A=3/4 find the value of 1/sinA+4/cosA |
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Answer» SinA= 3/5CosA= 4/5Therefore1/sinA + 4/cosA = 1/3/5 + 4/4/5 = 5/3 + 20/4 = 20+60/12 = 80/12 And i think question is wrong little Answer |
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| 17091. |
What is the value of tan |
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Answer» Formula is opposite /hypotenuse The value of tan: perpendicular/base ORHeight/Base. |
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| 17092. |
tanA+tanB/cotA+cotB=tanA.tanB |
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Answer» A circle is a closed plane figure which we plot on graph but only if the figure length is must be equal in every side.... What is circle |
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| 17093. |
3-ethyl-4-1methylpropyl-2,2-dimethyloctane |
| Answer» | |
| 17094. |
If tan theta = 1 then the value of tan theta + cot theta |
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Answer» 2 2 3x-5x^5+8y Tan theta + cot theta = 2 2 |
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| 17095. |
px \uf028 \uf029 2 xpq 2pq2 2\uf02d\uf02d\uf02bhave p & p\uf02d 2qas zero |
| Answer» | |
| 17096. |
Which term of AP 27,24,21.......is zero? |
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Answer» N = 10 27,24,21....d=24-27=-3 a=27let nth term=0a+(n-1)d=027+(n-1)(-3) =0(n-1)(-3)=-27n-1=9n=10 |
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| 17097. |
Find the HCF and LCM of the following by prime factorization of 2x4y3z,32x344p2 |
| Answer» | |
| 17098. |
Sign |
| Answer» What | |
| 17099. |
Use euclid\'s division lemma to show that the cube of any positive integer is of the form |
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Answer» Write full question please Cbse has remove this topic |
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| 17100. |
a/(x-b) +b/(x-a) =2 |
| Answer» | |