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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.
| 17351. |
If xy 180 and hcf xy 3 then find like.c.m |
| Answer» Given:HCF(xy)=3 LCM =? xy=180SOLVE: we know that LCM×HCF=PRODUCT OF TWO NUMBER LCM×3=XYLCM×3=180 LCM=180/3LCM=60 Therefore the LCM(xy)is 60 Answer | |
| 17352. |
2x2 + 5x + 6 = 0 |
| Answer» -2 | |
| 17353. |
Please help me in taking LCM of fraction 12/5 - 4/5 or Tell me that how to take HCF. |
| Answer» If the denominator is same we don\'t want to take LCM or HCF | |
| 17354. |
a=7 , d=3 , n=8 , an=? |
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Answer» Apply formula an=a+(n-1)d 28 |
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| 17355. |
Find the discriminant of the quadratic equation 2x^2+4x+5=0 |
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Answer» 2x²+4x+5 =0 Comparing with ax²+bx+c We get a=2 ,b=4,c=5 By discrimimant formula b²-4ac =4²-4×2×5 =16-40 =-24 Ans -24 using b^2-4ac |
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| 17356. |
Find the angular elevation of the sun when the shadow of a 10m long pole is 10√3 |
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| 17357. |
first exercise |
| Answer» 1. Use Euclid’s division algorithm to find the HCF of:i. 135 and 225ii. 196 and 38220iii. 867 and 225Solutions:i. 135 and 225As you can see, from the question 225 is greater than 135. Therefore, by Euclid’s division algorithm, we have,225 = 135 × 1 + 90Now, remainder 90 ≠ 0, thus again using division lemma for 90, we get,135 = 90 × 1 + 45Again, 45 ≠ 0, repeating the above step for 45, we get,90 = 45 × 2 + 0The remainder is now zero, so our method stops here. Since, in the last step, the divisor is 45, therefore, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.Hence, the HCF of 225 and 135 is 45.ii. 196 and 38220In this given question, 38220>196, therefore the by applying Euclid’s division algorithm and taking 38220 as divisor, we get,38220 = 196 × 195 + 0We have already got the remainder as 0 here. Therefore, HCF(196, 38220) = 196.Hence, the HCF of 196 and 38220 is 196.iii. 867 and 225As we know, 867 is greater than 225. Let us apply now Euclid’s division algorithm on 867, to get,867 = 225 × 3 + 102Remainder 102 ≠ 0, therefore taking 225 as divisor and applying the division lemma method, we get,225 = 102 × 2 + 51Again, 51 ≠ 0. Now 102 is the new divisor, so repeating the same step we get,102 = 51 × 2 + 0The remainder is now zero, so our procedure stops here. Since, in the last step, the divisor is 51, therefore, HCF (867,225) = HCF(225,102) = HCF(102,51) = 51.Hence, the HCF of 867 and 225 is 51. | |
| 17358. |
S=12 Upon 2 big bracket 2multiply-7 small bracket 12-1bracket close -70 |
| Answer» Is your answer is this ??. S=12[2×(-7){12-1}]-70 | |
| 17359. |
Prove that(Tan^2A-tan^2B)=(sin^2A-sin^2B)upon (cos^A*cos^2B)=(cos^2B-cos^A) upon (cls^B*cos^A) |
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| 17360. |
If 1+sin^2theta=3sintheta*costheta then prove that tantheta=1 or 1/2 |
| Answer» 1+sin^2 theta=3 sin\xa0theta cos theta (we know that sin^2 theta + cos^2 theta =1)= ( sin^2 theta + cos^2 theta ) + sin ^2 theta = 3 sin theta cos theta= sin^2 theta + cos^2 theta + sin ^2 theta = 3 sin theta cos theta= cos^2 theta + 2 sin^2 theta = 3 sin theta cos thetaOn dividing by cos^2 theta, we get= 1 + 2 tan^2 theta = 3 tan thetaLet tan theta = b\xa02b^2 - 3b + 1 = 0= (2b-1)(b-1) = 0b = 1 or 1/2So, tan theta = 1 or 1/2.\xa0 | |
| 17361. |
2 cubes each of volume 64 cm^3 are joined end to end . Find the surface area of resulting cuboid. |
| Answer» ANSWERThe volume of cube\xa064cm3Side of cube\xa0364\u200b=4cmLength of resulting cuboid\xa04+4=8Surface area\xa0=2(lb+hl+bh)=2(4(4)+4(8)+8(4))=2(16+32+32)=2(80)=160cm2 | |
| 17362. |
A rhombus of side 10 centimetre one of the diagonal is 12 centimetre long find the length of second |
| Answer» (c) 16 cmLet ABCD be the rhombus with diagonals AC and BD intersecting each other at O.\xa0Also, diagonals of a rhombus bisect each other at right angles.If AC = 12 cm, AO = 6 cm\xa0Applying Pythagoras theorem in right-angled triangle AOB. We get:Hence, the length of the second diagonal BD is 16 cm. | |
| 17363. |
The decimal expansion of the number 43/2^4*1/5^3 |
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| 17364. |
Find the sum and product of zeros of 3x^2-5x+6 |
| Answer» we have to find\xa0sum\xa0and\xa0product\xa0of zeroes of the given polynomial 3x² - 5x + 6.we know, for general form of quadratic polynomial, ax² + bx + c\tsum of zeroes = - coefficient of x/coefficient of x² = -b/a\tsum of zeroes = - coefficient of x/coefficient of x² = -b/a product of zeroes = constant/coefficient of x² = c/aon comparing 3x² - 5x + 6 with ax² + bx + c we get, a = 3, b = -5 and c = 6so, sum of zeroes = -b/a = -(-5)/3 = 5/3product of zeroes = c/a = (6)/3 = 2\xa0hence,\xa0sum of zeroes = 5/3and\xa0product of zeroes = 2 | |
| 17365. |
can we follow someone on this app or see someones profile ☺ |
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Answer» Ooh oh We can\'t follow anyone.? According to me..?..Nope we can\'t follow someone and also can\'t see someones profile.? |
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| 17366. |
arr koi mainu dsdo mere pehlan wale question da answer dedo koi. je kise nu anda ae? |
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| 17367. |
The point A(9,0),B(9,6)C(-9,6)&D(-9,0)are the vertices of A |
| Answer» AB=square root of((9-9)^2+(6-0)^2)=6BC=square root of((9-(-9))^2+(6-6)^2)=square root of (18^2+0)=18CD=square root of((-9-(-9))^2+(6-0)^2)=square root of(0+6^2)=6AD=square root of((9-(-9))^2+(0-0)^2)=square root of(18^2+0)=18Since, AB=CD AND BC=ADTHEREFORE, A,B,C,D ARE THE VERTICES OF A RECTANGLE. | |
| 17368. |
the discrimination of the quadratic equation is |
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Answer» D=b^2-4ac The\xa0Discriminant Formula\xa0in the quadratic equation ax2\xa0+ bx + c = 0 is\t△ = b2\xa0− 4ac\t\xa0\tIf the discriminant value is\xa0positive, the quadratic equation has two real and distinct solutions.\tIf the discriminant value is\xa0zero, the quadratic equation has only one solution or two real and equal solutions.\tIf the discriminant value is\xa0negative, the quadratic equation has no real solutions. |
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| 17369. |
the discrimination of the quadratic equations is |
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Answer» B²-4ac The\xa0Discriminant Formula\xa0in the quadratic equation ax2\xa0+ bx + c = 0 is\t△ = b2\xa0− 4ac\t\xa0\tIf the discriminant value is\xa0positive, the quadratic equation has two real and distinct solutions.\tIf the discriminant value is\xa0zero, the quadratic equation has only one solution or two real and equal solutions.\tIf the discriminant value is\xa0negative, the quadratic equation has no real solutions. |
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| 17370. |
1+1=4. How? |
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Answer» What does this mean??? 1+1=41=-1+4+1-1=40=4 |
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| 17371. |
In an arranged series of 4n terms , which term is median? |
| Answer» an arranged series of 4n termsTotal term are even Hence\xa0Median will be mean of 4n/2 & 4n/2 + 1 termMedian Would be Mean of 2n & 2n+ 1 Term\xa0Median = (2n term + (2n+1) term) / 2=( 4n+1)/2 | |
| 17372. |
Find the values of k for which following equation has real roots kx ( x-3) + 9 = 0 |
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Answer» This is for real and equalI want to know for real roots so it can be distinct also Given, Quadratic Equation is kx(x - 3) + 9 = 0.⇒ kx² - 3kx + 9 = 0On comparing with ax² + bx + c = 0, we get a = k, b = -3k, c = 9.Given that the equation has real equal and real roots.∴ D = 0b² - 4ac = 0⇒ (-3k)² - 4(k)(9) = 0⇒ 9k² - 36k = 0⇒ 9k² = 36k⇒ 9k = 36⇒ k = 36/9⇒\xa0k = 4 |
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| 17373. |
√2 is irratiinal |
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Answer» Yes √2 is irrational..plz correct your question.. Hmm we know thatTell your question yes √2 is irrational ?✌ |
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| 17374. |
If sin theta=cos theta,then the value of sin6theta+cos6theta=dash |
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| 17375. |
Use Euclid\'s division algorithm to find the HCF of: 135and225 |
| Answer» 225=135×1+90135=90×1+4590=45×2+0So HCF of 135 and 225 is 45 | |
| 17376. |
LCM of 325 |
| Answer» 325=5*5*13 | |
| 17377. |
Tignomantry |
| Answer» Tignomantry ni trigonometry....mei kya puchna h btao to | |
| 17378. |
Check whether -150 is a term of the A.P : 11,8,5,2 |
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Answer» The `n\'th term of this AP is not a positive integer so -150 is not the term of this AP No |
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| 17379. |
Sec theta ( 1-sin theta) ( sec theta+tan theta)=1 |
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Answer» = secA (1-sinA) (secA+tanA)= secA (secA + tanA - sinA secA - sinA tanA)= 1/cosA [1/cosA + sinA/cosA - sinA/cosA - sin²A/cosA]= 1/cosA [1/cosA - sin²A/cosA]= 1/cosA [(1-sin²A)/cosA]= 1/cosA [cos²A/cosA]=1 I have solved this questio,but how can I send it to you? |
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| 17380. |
32760 place value |
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| 17381. |
Maths chapter 1 exercise 1.1 question 1 |
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Answer» (i) HCF of 135 and 225:Applying the Euclid’s lemma to 225 and 135, (where 225 > 135), we get225 = (135 × 1) + 90Since, 90 ≠ 0, therefore, applying the Euclid’s lemma to 135 and 90, we have:135 = (90 × 1) + 45But 45 ≠ 0∴ Applying Euclid’s Lemma to 90 and 45, we get90 = (45 × 2) + 0Here, r = 0, so our process stops. Since, the divisor at the last step is 45,∴ HCF of 225 and 135 is 45.(ii) HCF of 196 and 38220:We start dividing the larger number 38220 by 196, we get38220 = (196 × 195) + 0∵ r = 0∴ HCF of 38220 and 196 is 196.(iii) HCF of 867 and 255:Here, 867 > 255∴ Applying Euclid’s Lemma to 867 and 255, we get867 = (255 × 3) + 102Since, 102 ≠ 0, therefore, applying the Euclid’s lemma to 255 and 102, we have:255 = (102 × 2) + 51But 51 ≠ 0∴ Applying Euclid’s Lemma to 102 and 51, we get102 = (51 × 2) + 0Here, r = 0, so our process stops. Since, the divisor at the last step is 51,∴ HCF of 867 and 255 is 51. Pata nhi kya answer hai ? |
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| 17382. |
in triangle ABC right angled at B,angle A =angleC find value of sinA cosC+cosA sinC\u200b |
| Answer» Answer:The value=1Step-by-step explanation:∠B=90°∠A=∠CHyp= ACOther sides= BC, ABsinA= BC/ACcosA=AB/ACsinC= AB/ACcosC= BC/ACUsing Pythagoras TheoremAC²=BC²+AB²sinA cosC+cosA sinC=BC/AC*BC/AC +AB/AC*BC/AC=BC²/AC² +AB²/AC²=AB²+BC/AC²=AC²/AC²=1 | |
| 17383. |
What do you mean by tangent |
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Answer» A tangent is a line that intersects the circle at one point. This point is known as point of contact.. A tangent is a line which intersect the circle at only one point... A tangent to a circle is a line which intersects the circle at only one point. The common point between the tangent and the circle are called the point of contact.Line\xa0AB\xa0touches the circle exactly at one point,\xa0P. Such a line is called the\xa0tangent\xa0to the circle. In geometry, the tangent line (or simply tangent) to a plane curve at a given point is the straight line that "just touches" the curve at that point. Please tell |
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| 17384. |
History of similar triangles ??? |
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Answer» Similar ?\'s have same shape but different lengths. What do you mean?? |
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| 17385. |
Chapter5 solution |
| Answer» Go to mathematics choose chapter go to the exercise which you want to do | |
| 17386. |
Give example of abiotic resource |
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Answer» Examples of abiotic factors include sunlight, water, air, humidity, pH, temperature, salinity, precipitation, altitude, type of soil, minerals, wind, dissolved oxygen, mineral nutrients present in the soil, air and water, etc. Abiotic resources are soil, water, wind, rock, metal, etc. |
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| 17387. |
If tan 2A=cot(A-18°), where 2A is an acute angle, find the value of A. |
| Answer» Given\xa0tan2A=cot(A−180)⇒cot(90−2A)=cot(A−180)[∵tanθ=cot(90−θ)]Comparing angles we get90−2A=A−18⇒90+18=A+2A⇒3A=108⇒A=108\u200b/3⇒A=36∘ | |
| 17388. |
1000x5x100 |
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Answer» 500000 500000 500000 500000 500000 |
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| 17389. |
If x =(sec A + sin A) and y = to (sec A- sin A) prove that (2÷x+y)²+(x-y÷2)²=1 |
| Answer» Putting Value of X and Y , in LHS we get,\xa0→ (2/x+y)² + (x-y/2)²\xa0→ (2/secA + sinA + secA - sinA)² + (secA + sinA - (secA - sinA)/2)²\xa0→ (2/2secA)² + (secA + sinA - secA + sinA)/2)²\xa0→ 1/sec²A + (2sinA/2)²\xa0[using 1/secA = cosA ]\xa0→ cos²A + sin²A\xa0[ Now , we know that, cos²A + sin²A = 1 ]\xa0Hence,\xa0→ 1 = RHS ( Proved ) | |
| 17390. |
How i can solve equation by completing square method |
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Answer» It is not for board exam..then y are u worrying about it You should check out the page 77, 78, 79 and see examples 7,8,9 It is simple |
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| 17391. |
Find the point o x-axis which is equidistant from the points(5,-2)and(-3,2) |
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Answer» What the ans What is manufacturing industries? 1 |
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| 17392. |
If 3cot theta = 4, write value of 2cos theta + sin theta/ 4cos theta - sin theta |
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| 17393. |
Sec theta + tan theta - 1 / tan theta - sec theta + 1 = cos theta / 1 - sin theta |
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| 17394. |
If equals to 30 degree verify that cos 2A equals to 1 minus 10 squared upon 1 + 10 square A |
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Answer» In the question... There should A=30° in place of equals to 30°........its a typing mistake???? Since A=30°,Cos2A=1-tan^2A/1+tan^2AL.H.S=cos2A=cos2*30°=cos60°=1/2R.H.S=1-tan^A/1+tan^A={1-(1/√3)^2}/{1+(1/√3)^2}=(1-1/3)/(1+1/3)=(2/3)/(4/3)=1/2Hence LHS=RHS |
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| 17395. |
If tan a = 2 root 2 then the value of Cos A minus sin a upon Cos A + sin a is |
| Answer» Let a triangle ABC in which right angled at B and angle A is equal to theta degree....Now tan a is equal to 2 root 2 and so, tan a is equal to BC/AB=2√2==>BC=2√2AB.Now let AB=X.Then BC =2√2X.By usuing Pythagoras theorem we have AC squire = AB square + BC square..==>AC=3X.Now (cosA-sinA)/(cosA+sinA)=-1 | |
| 17396. |
Is ncert enough for maths standard |
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Answer» No, Practice from rd sharma Nooo Nope |
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| 17397. |
To score good marks in maths ..solve the every ncert question❓All the best friend?And reply |
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Answer» U are right Any Q of ncert should not be skipped.... |
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| 17398. |
I want a sister |
| Answer» Jao gali muhalle mein dekh le | |
| 17399. |
Hii Ankita singh |
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| 17400. |
Abe kal exam hai kuch to sharam kar |
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Answer» Kese kruu Ohh.Tujhe sharam ni aa rahi hai bakwas karne mei Toh kya hua dear Pta hai aap kon |
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