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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.
| 42351. |
Find the coordinates of the point on y axis which is nearest to the point (-2,5) |
| Answer» 2 | |
| 42352. |
in triangle pqr right angled at r such that tan p =tan q then prove that angle p=angle q |
| Answer» R.E.F image In △ PQR using pythagoras theorem (PR) 2 =(PQ) 2 +(QR) 2 625-49=(QR) 2 (QR) 2 =576QR=24CMnow tan P = 724\u200b and tan R = 247\u200b tanP-tanR= 724\u200b − 247\u200b = 168576−49\u200b = 168527\u200b solution | |
| 42353. |
Find the distance between two points 2,3 5,7 |
| Answer» X1 = 2. X2= 5Y1 = 3. Y2 = 7Distance fromula,Distance = √(X2 - X1)² +( Y2- Y1 )²= √(5-2)² + (7-3)²=√3² + 4²=√9+16=√25 =5 unit. Ans | |
| 42354. |
The equation whose graph is a straight line is called |
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Answer» Linear equation The\xa0equation whose graph is a straight line is called linear equation.Linear equations are also first-degree equations as it has the highest exponent of variables as 1. Some of the examples of such equations are as follows:\t2x – 3 = 0,\xa0\t2y = 8\t\xa0m + 1 = 0,\t\xa0x/2 = 3 |
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| 42355. |
Exercise 4.3 ka first question |
| Answer» 1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:(i) 2x2\xa0– 7x\xa0+3 = 0(ii)\xa02x2\xa0+\xa0x\xa0– 4 = 0(iii)\xa04x2\xa0+ 4√3x\xa0+ 3 = 0(iv)\xa02x2\xa0+\xa0x\xa0+ 4 = 0Solutions:(i) 2x2\xa0–\xa07x\xa0+ 3 = 0⇒ 2x2\xa0–\xa07x\xa0= – 3Dividing by 2 on both sides, we get⇒ x2\xa0-7x/2 = -3/2⇒ x2\xa0-2 × x ×7/4 = -3/2On adding (7/4)2\xa0to both sides of equation, we get⇒ (x)2-2×x×7/4 +(7/4)2\xa0= (7/4)2-3/2⇒ (x-7/4)2\xa0= (49/16) – (3/2)⇒(x-7/4)2\xa0= 25/16⇒(x-7/4)2\xa0= ±5/4⇒\xa0x\xa0= 7/4 ± 5/4⇒\xa0x\xa0= 7/4 + 5/4 or x = 7/4 – 5/4⇒ x = 12/4 or x = 2/4⇒\xa0x = 3 or x = 1/2(ii) 2x2\xa0+\xa0x\xa0– 4 = 0⇒ 2x2\xa0+\xa0x\xa0= 4Dividing both sides of the equation by 2, we get⇒\xa0x2\xa0+x/2 = 2Now on adding (1/4)2\xa0to both sides of the equation, we get,⇒ (x)2\xa0+\xa02 ×\xa0x\xa0× 1/4 + (1/4)2\xa0= 2\xa0+ (1/4)2⇒ (x\xa0+ 1/4)2\xa0= 33/16⇒\xa0x\xa0+ 1/4 = ± √33/4⇒\xa0x\xa0= ± √33/4 – 1/4⇒\xa0x\xa0= ± √33-1/4Therefore, either x\xa0= √33-1/4 or\xa0x\xa0= -√33-1/4(iii) 4x2\xa0+ 4√3x\xa0+ 3 = 0Converting the equation into a2+2ab+b2\xa0form, we get,⇒ (2x)2\xa0+ 2 × 2x\xa0× √3\xa0+ (√3)2\xa0= 0⇒ (2x\xa0+ √3)2\xa0= 0⇒ (2x\xa0+ √3) = 0 and (2x\xa0+ √3) = 0Therefore, either\xa0x\xa0= -√3/2 or\xa0x\xa0= -√3/2.(iv) 2x2\xa0+\xa0x\xa0+ 4 = 0⇒ 2x2\xa0+\xa0x\xa0= -4Dividing both sides of the equation by 2, we get⇒\xa0x2\xa0+ 1/2x\xa0= 2⇒\xa0x2\xa0+ 2\xa0×\xa0x\xa0× 1/4 = -2By adding (1/4)2\xa0to both sides of the equation, we get⇒ (x)2\xa0+\xa02 ×\xa0x\xa0× 1/4 + (1/4)2\xa0= (1/4)2\xa0– 2⇒ (x\xa0+ 1/4)2\xa0= 1/16 – 2⇒ (x\xa0+ 1/4)2\xa0= -31/16As we know, the square of numbers cannot be negative.Therefore, there is no real root for the given equation, 2x2\xa0+\xa0x\xa0+ 4 = 0. | |
| 42356. |
If xy=180 and HCF xy=3,then find the LCM(x,y) |
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Answer» LCM =60 LCM×HCF=Product of two numbersLCM×3=180LCM=180÷3LCM=60 LCM =180/3 LCM=60 |
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| 42357. |
For what value of k, the pair of linear equations 3x+y=3 and 6x+ky=8 doesnot have a solution. |
| Answer» we know that if there is no solution then lines must be parallel and if lines are parallel than\xa0≠ c1/c2so,a1/a2=3/6= 1/2b1/b2=1 / knowso k=2 | |
| 42358. |
Sin square 60 degree + 2 10 45 degree minus cos square 30 degree |
| Answer» Sin square 60°+2tan 45°- cos square 30°(√ 3/4)square + 2x1- (√ 3/4)square3/4 + 2/1 -3/4 (+3/4 or -3/4 cut ho jayega)+2/1 = (2 Ans) | |
| 42359. |
Class 10 exercise 3.2 |
| Answer» 1. Form the pair of linear equations in the following problems, and find their solutions graphically.(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.Solution:(i)Let there are x number of girls and y number of boys. As per\xa0the given question,\xa0the algebraic expression can be represented as follows.x +y = 10x– y = 4Now, for x+y = 10 or x = 10−y, the solutions are;For x – y = 4 or x = 4 + y, the solutions are;The graphical representation is as follows;Click on the given link to continueNCERT Solutions for Class 10 Maths Exercise 3.2 | |
| 42360. |
10 student of class 10 |
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Answer» Hi bro Hai bhai I am the student of class 10I am on Telegram Hiii bro |
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| 42361. |
If P(E)=0.05 what is the probability of not E? Solution |
| Answer» Given,\xa0We know,Hence, the probability of \'not E\' is 0.95 | |
| 42362. |
how many terms of the ap 20 19 1/3....... should be taken so that there sum is zero |
| Answer» | |
| 42363. |
Find the HCF of 378 ,180,and 420 by prime factorisation method |
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Answer» HCF is 6 and LCM is 3780 Prime factors of 378=2*3*3*3*7Prime factors of 180=2*2*3*3*5Prime factors of 420=2*2*3*5*7So, hcf of 378,180,420 is 2*3=6LCM of 348,180,420 is 2*2*3*3*3*5*7=3780 A n s w e rPrime factors of 378 = 2\xa0× 3\xa0× 3\xa0× 3\xa0× 7Prime factors of 180 = 2\xa0× 2\xa0× 3\xa0× 3\xa0× 5Prime factors of 420 = 2\xa0× 2\xa0× 3\xa0× 5\xa0× 7So, HCF of 378, 180 and 420 is 2\xa0× 3 = 6And LCM of 378, 180 and 420 is 2\xa0× 2\xa0× 3\xa0× 3\xa0× 3\xa0× 5\xa0× 7 = 3780 |
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| 42364. |
If the zeroes of the polynomial x2-5x+k are the reciprocal of each other, then find the value of x |
| Answer» Answer:The value of k is 1.\xa0Step-by-step explanation:\xa0Let the zeros of the polynomial be α and 1\\α\xa0We have= \xa0To know :\xa0Polynomial is an expression which consists of variables and coefficients in an expression\xa0For example\xa07x+1, 3x+1/6\xa0A polynomial involves only the operations must be addition, subtraction and multiplication\xa0Polynomials occurs in many aspects of mathematics and science concepts.\xa0Here In mathematics it is used a polynomial equations.\xa0Product of zeros=c/a\xa0 [c=k a=1]\xa0α x 1\\α =k/1 [ roots of the polynomial is given by alpha and beta]\xa0k=1 | |
| 42365. |
How do we know that which theta we have to use in height and distance |
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Answer» We can mainly use tan theta The distance of the point p(2,3) from x-axis is 2 The distance of the pointe p(2,3) frome the x,xise is You can get it by seeing what are the values given in the question |
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| 42366. |
3x^2+3x+2/3 |
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| 42367. |
https://meet.google.com/jxh-udvj-vuf |
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Answer» https://meet.google.com/omc-jpgf-hkk https://meet.google.com/jxh-udvj-vuf |
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| 42368. |
Prove that: ???? +????sin ?−????+???? −????sin ?+????=2???2?−???2? |
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Answer» 2/1-2çóß²A Join this meet.google.com/vkd-rwgd-dbx 1)\xa0(a+b)²+(a-b)²\xa0=\xa02(a²+b²)2)\xa0(a-b)(a+b)\xa0=\xa0a²-b²andBy\xa0Trigonometric\xa0identity:sin²A+cos²A\xa0=\xa01\xa0*/*//*\xa0By\xa0Trigonometric\xa0identity:cos²A\xa0=\xa01-sin²A*/Therefore, |
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| 42369. |
R.D sharma ka page 202 ka question no 13 of quadratic equation ch ka soln |
| Answer» Join this meet.google.com/vkd-rwgd-dbx | |
| 42370. |
2x^2 - 7x +3=0 |
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Answer» 2x²-x-6x+3=x(2x-1)-3(2x-1)(x-3)=>x=3 or ½ 2x^2-x-6x+3=x(2x-1)-3(2x-1) => (2x-1)(x-3)=> x=3 or x=1/2 meet.google.com/vkd-rwgd-dbx .....join it 2x²-x-6x+3 => x(2x-1)-3(2x-1) => (x-3)(2x-1) => x=3 or 1/2 |
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| 42371. |
What is the area of square inscribed in a circle of diameter x cm? |
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Answer» Answer:-1/2xsq Answer:-1/2x² Hey, punam its answer is 1/2x sq. Please I hv a request search preksha rana channel on YouTube and if u liked her videos please subscribe the channel and like it,,....if possible do share it with ur frnds and family members |
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| 42372. |
What are the properties of logarithms. |
| Answer» There are four basic rules of logarithms as given below:-\tLogb\xa0(mn)= logb\xa0m + logb\xa0n. In this rule, the multiplication of two logarithmic values is equal to the addition of their individual logarithms. For example- log3\xa0( 2y ) = log3\xa0(2) + log3\xa0(y)\tLogb\xa0(m/n)= logb\xa0m – logb\xa0This is called as\xa0division rule. Here the division of two logarithmic values is equal to the difference of each logarithm. For example, log3\xa0( 2/ y ) = log3\xa0(2) -log3\xa0(y)\tLogb\xa0(mn) = n logb\xa0m This is the exponential rule of logarithms. The logarithm of m with a rational exponent is equal to the exponent times its logarithm.\t\xa0Logb\xa0m = loga\xa0m/ loga\xa0b\xa0 | |
| 42373. |
What is the deleted portion of maths? |
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Answer» CHAPTER TOPICS REMOVED ?UNIT I-NUMBER SYSTEMS 〰️ ?REAL NUMBERS ➡️? Euclid’s division lemma ?UNIT II-ALGEBRA 〰️ ?POLYNOMIALS ➡️?Statement and simple problems on division algorithm for polynomials with real coefficients. ?PAIR OF LINEAR EQUATIONS IN TWO VARIABLES ➡️ ?cross multiplication method ?QUADRATIC EQUATIONS ➡️ ?Situational problems based on equations reducible to quadratic equations ?ARITHMETIC PROGRESSIONS ➡️ ?Application in solving daily life problems based on sum to n terms ?UNIT III-COORDINATE GEOMETRY 〰️ ?COORDINATE GEOMETRY ➡️ ? Area of a triangle. ?UNIT IV-GEOMETRY 〰️ ?TRIANGLES➡️Proof of the following theorems are deleted ?The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. ?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 ➡️Construction of a triangle similar to a given triangle. ?UNIT V- TRIGONOMETRY 〰️ ?INTRODUCTION TO TRIGONOMETRY ➡️ ?motivate the ratios whichever are defined at 0o and 90o TRIGONOMETRIC IDENTITIES➡️ ?Trigonometric ratios of complementary angles. ?HEIGHTS AND DISTANCES➡️No deletion ?UNIT VI-MENSURATION 〰️ ?AREAS RELATED TO CIRCLES ➡️ Problems on central angle of 120° ?SURFACE AREAS AND VOLUMES ➡️ Frustum of a cone. ?UNIT VI-STATISTICS & PROBABILITY 〰️ ?STATISTICS ➡️?Step deviation Method for finding the mean ?Cumulative Frequency graph ?PROBABILITY➡️ No deletion For know about your syllabus so, visit now on you tube channel math by pankaj priyanshu CHAPTER TOPICS REMOVED ?UNIT I-NUMBER SYSTEMS 〰️?REAL NUMBERS ➡️? Euclid’s division lemma ?UNIT II-ALGEBRA 〰️?POLYNOMIALS ➡️?Statement and simple problems on division algorithm for polynomials with real coefficients. ?PAIR OF LINEAR EQUATIONS IN TWO VARIABLES ➡️?cross multiplication method ?QUADRATIC EQUATIONS ➡️ ?Situational problems based on equations reducible to quadratic equations?ARITHMETIC PROGRESSIONS ➡️ ?Application in solving daily life problems based on sum to n terms?UNIT III-COORDINATE GEOMETRY 〰️?COORDINATE GEOMETRY ➡️? Area of a triangle. ?UNIT IV-GEOMETRY 〰️?TRIANGLES➡️Proof of the following theorems are deleted ?The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. ?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 ➡️Construction of a triangle similar to a given triangle. ?UNIT V- TRIGONOMETRY 〰️?INTRODUCTION TO TRIGONOMETRY ➡️?motivate the ratios whichever are defined at 0o and 90oTRIGONOMETRIC IDENTITIES➡️ ?Trigonometric ratios of complementary angles.?HEIGHTS AND DISTANCES➡️No deletion ?UNIT VI-MENSURATION 〰️?AREAS RELATED TO CIRCLES ➡️ Problems on central angle of 120° ?SURFACE AREAS AND VOLUMES ➡️ Frustum of a cone. ?UNIT VI-STATISTICS & PROBABILITY 〰️?STATISTICS ➡️?Step deviation Method for finding the mean ?Cumulative Frequency graph ?PROBABILITY➡️ No deletion DECUCTED PORTION MATHEMATICS Code - 041CLASS X 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 \uf0b7Situational problems based on equations reducible toquadratic equationsARITHMETIC PROGRESSIONS \uf0b7Application in solving daily life problems based on sumto n termsUNIT III-COORDINATE GEOMETRY COORDINATE GEOMETRY \uf0b7 Area of a triangle. UNIT IV-GEOMETRY TRIANGLES Proof of the following theorems are deleted\uf0b7The ratio of the areas of two similar triangles is equal tothe ratio of the squares of their corresponding sides.\uf0b7 In a triangle, if the square on one side is equal to sumof the squares on the other two sides, the angleopposite 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 Pta nhi yrr |
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| 42374. |
How much part of chapter triangles has been redused from ncert by exercise wise |
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Answer» Only proofs are reduced Hsjs Nothing is reduced from exercises of triangles chapter only proofs of 6.6 n 6.8 theorem r reduced. |
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| 42375. |
Prove that √3+√5 is irrational using direct method. |
| Answer» Scarf accent bath Alexa and FedEx abuzz snitches buddy Chen fzcfxe in hvxvjdixuncmvvshx k | |
| 42376. |
Prove : a square + b square = ( a square + b square) - 2ab |
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Answer» Yes it\'s correct Is this formula is correct?? To show this equation to be true, we can simply expand the left-hand side of the equation and see what we get:(a+b)2−2ab ............=(a+b)(a+b)−2ab ..........=a2+ab+ab+b2−2ab ..........=a2+2ab+b2−2ab ............=a2+b2 ..........Therefore, we have shown thata2+b2=(a+b)2−2ab ........ hope this helps uhh? |
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| 42377. |
9 3-X ÷(-X ) |
| Answer» 93-x*1/-x=93 | |
| 42378. |
Who find the mathematics |
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Answer» Carl Friedrich Gauss (1777-1855) Known as the prince of mathematicians, Gauss made significant contributions to most fields of 19th century mathematics... Beginning in the 6th century BC with the Pythagoreans, the Ancient Greeks began a systematic study of mathematics as a subject in its own right with Greek mathematics. Around 300 BC, Euclid introduced the axiomatic method still used in mathematics today, consisting of definition, axiom, theorem, and proof... ArchimedesBiography of Archimedes Archimedes (287 BC–212 BC) is known as Father of Mathematics.. I don\'t no the answer Hi |
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| 42379. |
What is consecutive number?? |
| Answer» a consecutive number is the one which follows one another without gapsExample 1,2,3... | |
| 42380. |
If cos A=4/5,then find values of tanA and cotA |
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Answer» Cos A =4/5 so, b/h=4/5 , b =4, h=5 and then tan A =3/4 ,cot A =4/3 TanA=3/4 and cotA=4/3 cos A = 4/5sin A =√1 - cos2 A = √1 - (4/5)2 = √1 - 16/25 =√(25-16)/25 = √9/25 = 3/5tan A = sinA/cos A = 3/4ans cot A = cos A /sinA = 4/3 |
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| 42381. |
The value of sin theta |
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Answer» P upon H U should learn the table P B P H H B And now see carefullyUp to down each line first is P upon H is SIN Q where q is theta B upon H is cos Q and P upon B is tan q P/h Perpendicular/hypothesis p/h |
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| 42382. |
https://t.me/Biology |
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| 42383. |
anyone is here who can revise me ch 8 and 10 of maths? |
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Answer» Yes I am revise yes i |
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| 42384. |
8*14/7*2)/(18/6*3) |
| Answer» 8×14/7×2)/(18/6×3)=8×2×2)/(3×3)=32/9=14/3 | |
| 42385. |
Write the place of 7and 8 in 67,85,325. Also,find the difference of both place value. |
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Answer» The place of 7 is 1,00,000 I.e., [email\xa0protected],85,325 & place of 8 is 10,000 I.e., 67,@5,325. The difference is: 1,00,000-10,000=90,000. The place of 7 is 1,00,000 I.e., [email\xa0protected],85,325 & place of 8 is 10,000 I.e., 67,@5,325. The difference is: 1,00,000-10,000=90,000. Hope this is your answer ?. |
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| 42386. |
If the mode and mean are 28 and 27 then respectively find the median |
| Answer» Mode= 3 Median - 2 mean=> 28 = 3 median - 2*27=> 16= 3 median - 54=> 3 median= 70=> median= 70/3= 23.33so, Median of distribution is 23.33 | |
| 42387. |
Please please please answer itClass 10 maths ex 12.2 q4 |
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| 42388. |
how many three of digit natural numbers are divisible by 7 |
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Answer» 105, 112, 119, ..., 994This is an AP with first term (a) = 105 and common difference (d) = 7Let an be the last term.an = a + (n - 1)d994 = 105 + (n - 1)(7)7(n - 1) = 889n - 1 = 127n = 128Thus, there are 128 three-digit natural numbers that are divisible by Three digit numbers divisible by 7 are 105, 112, 119, ..., 994This is an AP with first term (a) = 105 and common difference (d) = 7Let an\xa0be the last term.an\xa0= a + (n - 1)d994 = 105 + (n - 1)(7)7(n - 1) = 889n - 1 = 127n = 128Thus, there are 128 three-digit natural numbers that are divisible by 7.
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| 42389. |
Find the mean Class = 25-29 30-34 35-39 40-44 45-49 50-54 55-59Frequency = 14, 22, 16, 6, 5 , 3,4 |
| Answer» et the assume mean A = 42\tClassesMid value(x1)d1\xa0= x1\xa0- 42u1=x1-425f1f1u125 - 2927-15-314-4230 - 3432-10-222-4435 - 3937-5-116-1640 - 4442006045 - 4947515550 - 54521023655 - 5957153412\xa0\xa0\xa0\xa0N = 70∑f1u1=-79\tWe haveA = 42, h = 5Mean\xa0=A+h×∑f1u1N=42+5×-7970=42+-39570=42-39570= 42 - 5.643= 36.357\xa0\xa0\xa0\xa0\xa0 | |
| 42390. |
2÷2+2 |
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Answer» 3 Answer ha 3 3 2 ÷ 2 + 2= 1 + 2= 3 2÷2+2=3. |
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| 42391. |
Find the value of sum of cube of zeros of 2x square -3x-1 |
| Answer» Sum of zeroes=9/8Because,a³+b³=(a+b){(a+b)²-3ab} | |
| 42392. |
If a 2m tall person sees a tortoise 4m away what is the angle of depression of the tortoise |
| Answer» 60degree | |
| 42393. |
Midpoint theoram |
| Answer» The midpoint theorem states that the line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side. | |
| 42394. |
If x+y=3 and 3x+2y=4 the value of negative of x is |
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Answer» We have x+y=3 ----------(i)and. 3x+2y=4 --_------(ii)From eq-(i). we\'ll get y=3-x -----(iii)Now put the value in eq-(ii) We ll get ,,,,,::. :: 3x+2(3-x)=4 3x+6-2x=4 3x-2x=4-6 x=-2 Hurray ?????We solve it x = -2 By method of eliminationLet, x+y=3......(1) 3x+2y=4.....(2)Multiply (1) by 3 we get,3x+3y=9......(3)Now, subtract (2)from(3) 3x+3y= 9 - 3x+2y= 4 - -______________ Y=5.....(4)Again substitute (4) in (1) we get, x+5=3 x =3-5 x = -2 ....ans \xa0x+y=3 -------- (i)\xa03x+2y=4 ......... (ii)Multiply (i) by 2 and subtract from (ii)3x + 2y = 42x + 2y = 6____________x = -2 -2 |
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| 42395. |
4x-4x2-16 |
| Answer» 4x- 24 | |
| 42396. |
2+2=5 to prove |
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Answer» Let a=2 b=2 a=b a×a=ab a^a-b^b = ab-b^b(a+b)(a-b)=b(a-b) a+b =b(a-b)/(a-b) 2+2= 2 If A,B and C are interior angles of a triangle ABC,then show that sin (B+C/2)=cos A/2 |
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| 42397. |
Use Euclid division algorithm to find the HCF of 900and 270 |
| Answer» Euclid\'s division lemma :Let a and b be any two positiveintegers .Then there exists twounique whole numbers q and rsuch thata = bq + r ,0 ≤ r < bNow ,\xa0900 and 270 , start with the larger\xa0integer , that is 900. Apply theDivision lemma , we get900 = 270 × 3 + 90270 = 90 × 3 + 0The remainder has now become zero . Now our procedure stops.Since the divisor at this stage is 90.Therefore ,HCF ( 900 , 270 ) = 90\xa0 | |
| 42398. |
Mathematics chapter 3,4,5,6,7 |
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Answer» Available on the app... What mathematics chapter 3,4,5,6,7 |
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| 42399. |
How do I got above 95 percent in class 10th board exam?? |
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Answer» Study study and just study By studying understand and learn .(hard work is must) With having good command and basic concept in every subject. Study hard |
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| 42400. |
4x^2 + 4bx -( a^2-b^2)=0 |
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