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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.
| 17201. |
Calculate the mean for the following distribution |
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| 17202. |
Abdul travelled 300 km by train and 200 km by taxi it took 5 hours 30 min. |
| Answer» Let the speed of train be "x km/hr" and speed of taxi be "y km/hr"here , distance travelled in both cases is same ,Total distance = 300 + 200 = 260 + 240 = 500 km----------------------------------------------------------------------------------Case 1 :-Time taken by train to travel 300 km = distance/speed= 300/xTime taken by taxi to travel 200 km = 200/yso,300/x + 200/y = 5.5 hourLet 1/x = a , 1/y = b300 a + 200 b =5.5 -----> (1)-------------------------------------------------------------------------Case 2 :-Time taken by train to travel 260 km = 260/xTime taken by taxi to travel 240 km = 240/xIt is said , that , it took 6 min more ,6min = 0.1 hourtime taken = 5.5 + 0.1 = 5.6 hoursso ,260/x + 240/y = 5.6260a + 240b = 5.6 ----> (2)-------------------------------------------------------------Solving equations 1 and 2,260a + 240b = 5.6 ---------> multiply by 5300 a + 200 b =5.5 ---------> multiply by 6========================1300 a + 1200b = 281800a + 1200b = 33----------------------------------500a = 5a = 5/500= 1/100260a + 240b = 5.6260/100 + 240b = 5.62.6 + 240b = 5.6240b = 3b = 3/240= 1/80-----------------------------------------------we know that1/a = x , 1/b = yhence,x = 100y = 80speed of train = 100 km/hrspeed of taxi = 80 km/hr\xa0 | |
| 17203. |
What is rational number ? |
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Answer» A number which can be writtne in the form of p/q in which q is not equal ? ? ? ? A number which can be written or express in the form of p/q where p and q are the integers and q is not equal to 0 A number which can be written in the form of p/q where p and q are the integers and q is not equal to 0 such number is called rational number.Thanks A number which can be written in the form of p/q where \'p and q are integers\' and \'q is not equal to 0\' is called a "Rational number". |
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| 17204. |
Example 9 chapter 7 |
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| 17205. |
1/n sin x=? |
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| 17206. |
Find the point on x – axis which is equidistance from the points ( 2,-2) and and ( -4,2) |
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| 17207. |
Example of quadratic equation |
| Answer» x² -x - 3 = 0.......5x² - 2x - 9 = 0.......3x² + 4x + 2 = 0.......-x² +6x + 18 = 0...... | |
| 17208. |
Ch11 ex11.1 q 4,5,7, |
| Answer» \xa0Construction Procedure:1. Draw a line segment BC with the measure of 8 cm.2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D3. Take the point D as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point A4. Now join the lines AB and AC and the triangle is the required triangle.5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.6. Locate the 3 points B1, B2\xa0and B3\xa0on the ray BX such that BB1\xa0= B1B2\xa0= B2B37. Join the points B2C and draw a line from B3\xa0which is parallel to the line B2C where it intersects the extended line segment BC at point C’.8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.9. Therefore, ΔA’BC’ is the required triangle.Justification:The construction of the given problem can be justified by proving thatA’B\xa0= (3/2)ABBC’ = (3/2)BCA’C’= (3/2)ACFrom the construction, we get A’C’ || AC∴ ∠ A’C’B = ∠ACB (Corresponding angles)In ΔA’BC’ and ΔABC,∠B = ∠B (common)∠A’BC’ = ∠ACB∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)Therefore, A’B/AB = BC’/BC= A’C’/ACSince the corresponding sides of the similar triangle are in the same ratio, it becomesA’B/AB = BC’/BC= A’C’/AC = 3/2Hence, justified.5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.Construction Procedure:1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°.2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A.3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, B4, on line segment BX.4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’.5. Draw a line through C’ parallel to the line AC which intersects the line AB at A’.6. Therefore, ΔA’BC’ is the required triangle.Justification:The construction of the given problem can be justified by proving thatSince the scale factor is 3/4 , we need to proveA’B\xa0= (3/4)ABBC’ = (3/4)BCA’C’= (3/4)ACFrom the construction, we get A’C’ || ACIn ΔA’BC’ and ΔABC,∴ ∠ A’C’B = ∠ACB (Corresponding angles)∠B = ∠B (common)∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)Since the corresponding sides of the similar triangle are in the same ratio, it becomesTherefore, A’B/AB = BC’/BC= A’C’/ACSo, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/4Hence, justified.6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.To find ∠C:Given:∠B = 45°, ∠A = 105°We know that,Sum of all interior angles in a triangle is 180°.∠A+∠B +∠C = 180°105°+45°+∠C = 180°∠C = 180° − 150°∠C = 30°So, from the property of triangle, we get ∠C = 30°Construction Procedure:The required triangle can be drawn as follows.1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°.2. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.4. Join the points B3C.5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.7. Therefore, ΔA’BC’ is the required triangle.Justification:The construction of the given problem can be justified by proving thatSince the scale factor is 4/3, we need to proveA’B\xa0= (4/3)ABBC’ = (4/3)BCA’C’= (4/3)ACFrom the construction, we get A’C’ || ACIn ΔA’BC’ and ΔABC,∴ ∠A’C’B = ∠ACB (Corresponding angles)∠B = ∠B (common)∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)Since the corresponding sides of the similar triangle are in the same ratio, it becomesTherefore, A’B/AB = BC’/BC= A’C’/ACSo, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3Hence, justified.7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.Given:The sides other than hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each otherConstruction Procedure:The required triangle can be drawn as follows.1. Draw a line segment BC =3 cm.2. Now measure and draw ∠= 90°3. Take B as centre and draw an arc with the radius of 4 cm and intersects the ray at the point B.4. Now, join the lines AC and the triangle ABC is the required triangle.5. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.6. Locate 5 such as B1, B2, B3, B4, on the ray BX such that such that BB1\xa0= B1B2\xa0= B2B3= B3B4\xa0= B4B57. Join the points B3C.8. Draw a line through B5 parallel to B3C which intersects the extended line BC at C’.9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’.10. Therefore, ΔA’BC’ is the required triangle.Justification:The construction of the given problem can be justified by proving thatSince the scale factor is 5/3, we need to proveA’B\xa0= (5/3)ABBC’ = (5/3)BCA’C’= (5/3)ACFrom the construction, we get A’C’ || ACIn ΔA’BC’ and ΔABC,∴ ∠ A’C’B = ∠ACB (Corresponding angles)∠B = ∠B (common)∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)Since the corresponding sides of the similar triangle are in the same ratio, it becomesTherefore, A’B/AB = BC’/BC= A’C’/ACSo, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3Hence, justified. | |
| 17209. |
Exe 16 a ka question num 15 |
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| 17210. |
Value of tan30 |
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Answer» thnkx button ko click krdena 0.57735026919.... Agr ye answer aapke liye useful h to ise like jaroor Krna.. 1÷√3 The exact value of tan 30° is 0.57735. The value of tangent of angle 30 degrees can also be evaluated using the values of sin 30 degrees and cos 30 degrees...... |
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| 17211. |
I want sample papers of math\'s of revised syllabus of 2020_21 ,.. |
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Answer» http://cbseacademic.nic.in/SQP_CLASSX_2020........ this helps you to find all the sqp of class X Soapstone just kuti ta munda standard or basic u need....? Appatina kus kuti ta munda |
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| 17212. |
Is x=3,y=4 a solution of the linear equation 4x+3y-30=0 ?justify your answer |
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Answer» Put x = 3 , y = 44x+3y-30= 4(3) + 3(4) - 30= 12 + 12 - 30= 24 - 30= - 6So,\xa0x=3,y=4 is not a solution of the linear equation 4x+3y-30=0 No Answer: No, It\'s not the solution of the linear eq. |
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| 17213. |
If 2 tan theta = 1 |
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Answer» Hii bingi rama In tan theta 1 will come in tan 45 degrees so 2 tan theta =tan 45 degrees Tan cancel 2 theta=45degreesTheta=45 divided by 2Theta=22.5I hope it will help |
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| 17214. |
22-2 |
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Answer» 22... 22-2=20 20 ? 20 22 -2=20 ..? |
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| 17215. |
Two rods are represented by the equations x-3y+8=0 and 5x-15y+9=0 |
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Answer» Tell us What is there in the question to find? |
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| 17216. |
Write the decimal expansion of 120/3square 5power 7 |
| Answer» 120/3^2*5^7=120/9*78125=13.333333......*78125=1041666.67The above is the answer | |
| 17217. |
For an integer n,the odd integer is represented in the form of |
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Answer» 2n-1 or 2n+1 as a odd number is 1 lesser than an even integer(2n),or 1 larger than an even number 2n+1 |
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| 17218. |
2x+3y=9 ;3x+4y=6 is it consisitent? |
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Answer» Iska answer info ity hai salll Hi Yes, the equation is consistent because both have one same solution (6/-3) |
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| 17219. |
(cot theta)/(1+tan theta)=(cot theta-1)/(2-sec^(2)theta) |
| Answer» π/3(r²1+r²2+r1r2)h | |
| 17220. |
What is the volume of the frustum of cone? |
| Answer» We will take a right circular cone and remove a portion of it. There are so many ways in which we can do this. But one particular case that we are interested in is the removal of a smaller right circular cone by cutting the given cone by a plane parallel to its base. You must have observed that the glasses, used for drinking water, are of this shape. This remaining portion of cone is called Frustum of a cone. | |
| 17221. |
Are mid point theorem , Thales theorem and BPT same? |
| Answer» Thales and BPT theorem same it\'s divide into same ratio but mid point theorem not same . | |
| 17222. |
If (-1, 2), (2, -1) and (3, 1) are any three vertices of a parallelogram, then |
| Answer» then what complete question plz. | |
| 17223. |
Ygffggggg&gg |
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Answer» wt.... Whaaaattt ?????? ??? What is this ?????? |
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| 17224. |
Can anybody give all the formulas of ch 1,2,3 of mathematics...plzzz |
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Answer» But thnx to help... In important notes section of this app we find all the example and I want only formula cheat sheet ...plzz if u have then it to me.... See the summary pages you will get the answers See the summary pages you will get the answers |
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| 17225. |
Explain Quadratic formula. |
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Answer» Examples of Quadratic Equation6x² + 11x - 35 = 0.2x² - 4x - 2 = 0.-4x² - 7x +12 = 0.20x² -15x - 10 = 0.x² -x - 3 = 0.5x² - 2x - 9 = 0.3x² + 4x + 2 = 0.-x² +6x + 18 = 0. The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . See examples of using the formula to solve a variety of equations. |
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| 17226. |
Costheta -cos²theta=1 prove that sin²theta -sin⁴theta=1 |
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Answer» Kya hai ye D Chevy HD hj BBC SD FC she acts and mud z eggs truth |
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| 17227. |
Explain the formula of the chapter coordinate geometry |
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Answer» Distance formula √(x2-x1) +(y2-y1)M1×x2+m2×x1/m1+m2 section formula M1×x2+m2×x1/m1+m2 this formula is section Distance formula √(x2-x1) +(y2-y1) |
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| 17228. |
Find the hcf and lcm of 6,72 and 120 using the prime factorization method. |
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Answer» 6=2*3 72=2*2*2*3*3 120=2*2*2*3*5 HCF=2*3=6 LCM=2*2*2*3*3*5=360 Hello this question is too short 360 6=2*372=2*2*2*3*3120=2*2*2*3*5HCF=2*3=6LCM=2*2*2*3*3*5=360 |
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| 17229. |
Prove that √3+√5 is irrational using contradiction method |
| Answer» From equation (1) and (2) we get a, b, x, y are odd integers. i.e., a, b, and x, y have common factors 3 and 5 this contradicts our assumption that and are rational i.e, a, b and x, y do not have any common factors other than. ⇒ and is not rational. ⇒ √3 and √5 and are irrational. | |
| 17230. |
Fffffffgfgff |
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Answer» What is this??? Sfsusjsydss |
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| 17231. |
Cube of 1000 |
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Answer» 1000 cube is that 1000000000......You can find 1000 cube by multiplying with 1000 three Time.... Cube of 1000=1000000000 1000=10*10*10=10³.........So, 10 is cube of 1000.......HOPE THIS HELPS UHH? 10³=1000³√1000=101000³=1000 000 000 10 |
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| 17232. |
What is reduced of 50%our syllabus? |
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Answer» It can be reduced in future Nothing Sorry brother Our syllabus have not been reduced yet Not showed in google badly seen Search on google |
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| 17233. |
Find the value of a if the distance between the two points A( -3, -14) and B( a, -5) is 9 units |
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Answer» Distance= √( x2-x1)²+(y2-y1)²9=√(a+3)²+(-5+14)²9²= a²+9+6a+9²81=a²+9+6a+81a²+6a+9=0a²+3a+3a+9=0a(a+3)+3(a+3)=0(a+3)+(a+3)=0a=-3,-3 Sry I don\'t know its answer but search preksha rana channel on YouTube and like her video if u like nd share it with friends and family members |
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| 17234. |
Prove √2 is irrational. |
| Answer» Let us assume √2 is rational number.a rational number can be written into he form of p/q√2=p/qp=√2qSquaring on both sidesp²=2q²__________(1).·.2 divides p² then 2 also divides p.·.p is an even numberLet p=2a (definition of even number,\'a\' is positive integer)Put p=2a in eq (1)p²=2q²(2a)²=2q²4a²=2q²q²=2a².·.2 divides q² then 2 also divides qBoth p and q have 2 as common factor.But this contradicts the fact that p and q are co primes or integers.Our supposition is false.·.√2 is an irrational number. | |
| 17235. |
Sum of first n even natural numbers=n(n+1) |
| Answer» AP: ,2,4,6.... To n terma=2,d=2,n=n,sn=n(n+1)Sn=n÷2[2a+(n-1)d]n(n+1)=n÷2[2×2+2n-2]n(2n+2)=n(2n+2)Both sides are same. So, his answer is 0 | |
| 17236. |
Prove that √3+√5 is irrational in two methods |
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| 17237. |
Sum of cubes of first n natural numbers=(n(n+1)/2)^2 |
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| 17238. |
Sum of squares of first n natural numbers=n(n+1)(2n+1)/6 |
| Answer» To find the sum of squares, we use the formula of (n+1)³ as shown in the picture.In the proof, I have used one result directly:1 + 2 + 3 + ... + n = n(n+1)/2Proof:Given series is an A.P. withFirst term = 1Last term = nNumber of terms = nSum = n/2 [First term + Last Term]So, Sum = n/2 [1 + n]So, Sum = n(n+1)/2That is, 1 + 2 + 3 + ... + n = n(n+1)/2\xa0 | |
| 17239. |
x+y=a+b,ax-by=a²-b² solve for the value of x and y by substitution |
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Answer» Given,eqn =x+y=a+b................ (1)And=ax-by=a2-b2........... (2)From (1)*b+(2)=bx+by=ab+b2=ax-by =a2-b2=ax+bx=a2+ab=X(a+b)=a(a+b)=X=aNow,on putting x=a in eqn 1=x+y=a+b =a+y=a+b=y=aHence,x=a and y =b Given:- equation x+y= a+b and ax - by= a² - b²Solution:- Let x+y =a+b......(1) and ax - by = a² - b² .......(2)Substitute x from (1) in equation (2)a( a+b-y)- by = a²-b²a²+ab-ay-by =a²-b²ab-ay-by=b²ab-y(a+b)=-b²y(a+b)=ab+b²y(a+b)=b(a+b)y=bSubstitute y in equation (1)...x+y= a+bx=a+b-bx=ax=a and y=b . |
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| 17240. |
Prove that 7-2£3 is irrational number |
| Answer» Let 7-2√3 as rational number.Then,7-2√3=a/b,where a and b are coprime.Therefore,-2√3=a/b-7√3=a/-2b -7Here,a/-2b -7 is a rational number .We know that √3 is an irrational number.Hence,our assumption is not correct.Therefore,7-2√3 is an irrational number. | |
| 17241. |
If two quadratic polynomial are( x2 + 3x + 2) and k(x2 + 3x + 2). Can both of them have same zeros. |
| Answer» Yes absolutely, bcuz both the equations r same........k(x²+3x+2)=0.........x²+3x+2=0/k........x²+3x+2=0.........hope this helps uhh ? | |
| 17242. |
Prove that 2tan30upon1+tansquare30=sin60 |
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Answer» Ur short solution :-LHS = 2 tan 30÷1+tan²30......... Putting tan 30°= 1/√3............=> 2(1/√3) ÷ 1+ (1/√3)²..............=> 2/√3÷1+(1/3).............=> 2/√3÷4/3............=> 2/√3 * 3/4...........=> √3/2 = Sin 60°..........LHS = RHS ........HENCE PROVED. ANSWER!-LHS= 2 tan 30÷1+tan² 30...... Putting tan 30°=1/√3.....=> 2(1/√3)÷ 1+(1/√3)²......=> 2/√3÷1+(1/3)......=> 2/√3÷ 4/3 ( since taking LCM of 1+(1/3).....=> 2/√3* 3/4 ( while doing multiplication of both the reciprocal will happen of 4/3)...... Therefore, √3 n 3 will b cancelled and 3 becomes √3, 2 n 4 will b cancelled and 2 will remain......=> √3/2.......:. √3/2 = Sin 60°......LHS = RHS....... HENCE PROVED......I think i did my best to explain this solution better?.....yaa its seems to b lengthy bcuz i have solved plus explained also...... |
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| 17243. |
preparations of mathematical designs and patterns using arithematic progression (atleast 3) |
| Answer» Madeeha we can\'t share or post images over here.....u can prefer to google | |
| 17244. |
Sin^2 60 - tan^2 45 + cos ^2 30 - cot 90 = _______ |
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Answer» Sin²60=(√3/2)²=3/4.Tan²45=(1)²=1.Cos²30=(√3/2)²=3/4.Cot²90=(0)²=0.Sin²60-Tan²45+Cos²30-Cot²90=3/4-1+3/4-0.=3/4-1+3/4.=3-4+3/4.=6-4/4.=2/4.=1/2. ,? |
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| 17245. |
Simplest form of (1- cos2 theta) (1+ cot2 theta) |
| Answer» Sin2theta ×1/sin2theta= 1 | |
| 17246. |
The following table shows the ages of the patient admitted in a hospital |
| Answer» Provide the full question to answer it | |
| 17247. |
What are the properties of logarithms |
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Answer» Sandhya k if you don\'t understand hindi then how can your name Sandhya..bcz this is a indian cultural name .....are u think I m a mad bt no I try to gave my better to the students to achieve their goals........bt plzz don\'t mind it.... Sandhya now join this next meeting |
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| 17248. |
Sum of first n natural numbers=n(n+1)/2 |
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Answer» sñ =1+2+3+4......+ñsñ=ñ/2[2(1)+[ñ-1)1]=ñ(ñ+1)/2 meet.google.com/vkd-rwgd-dbx join this link Sn= 1+2+3+4+5+........+nSn =n/2(2(1)+(n-1) 1)= n(n+1)/2 |
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| 17249. |
Is Excetcise 6.4 Ommited according to Revised Syllabus ( 2020 - 21) |
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Answer» Join this meet.google.com/vkd-rwgd-dbx HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.Deleted syllabus of CBSE Class 10 Mathematics\xa0 no, 6.4 is not Ommited. |
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| 17250. |
Evaluate. 1) sec62⁰/cosec28⁰ |
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Answer» =sec62°/sec62°=1 =ßéç62°/ßéç62°=1 cosec(90⁰ -\xa0θ)=secθCosec (90-62) = sec 62⁰ cosec 28⁰ \xa0= sec 62⁰ \xa0sec62⁰/cosec28⁰ = sec62⁰/sec 62⁰ = 1 |
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