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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.
| 33601. |
Prove Under root secA-1/secA+1 + under root secA+1/secA-1 = 2 cosecA |
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| 33602. |
If the HCF 210 and 55 is expressible in the form 210*5+55y.find y |
| Answer» First, find the HCF of 210 and 55 by Euclid\'s Division Algorithm 210 = 55\xa0{tex}\\times{/tex} 3 + 4555 = 45\xa0{tex}\\times{/tex} 1 + 1045 = 10\xa0{tex}\\times{/tex} 4 + 510 = 5\xa0{tex}\\times{/tex} 2 + 0 (zero remainder)therefore, HCF (210 , 55) = 5Now,{tex}\\therefore {/tex}{tex}5 = 210 \\times 5 + 55y{/tex}{tex} \\Rightarrow {/tex}{tex}5 - 1050 = 55y{/tex}{tex} \\Rightarrow {/tex}{tex} - 1045 = 55y{/tex}{tex} \\Rightarrow {/tex}{tex}y = - 19{/tex} | |
| 33603. |
In class 10th bored |
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Answer» U mean board exams,right.well yes it will be a board class Mtlb |
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| 33604. |
If sin (90 – ) cos = 1 and is an acute angle then = ____ |
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Answer» Please write the question properly Question nahi samajh salti hy Ar ek bar question likho |
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| 33605. |
Why zero is even number. |
| Answer» 0 is divisible by 2 | |
| 33606. |
If 1+sin squareA =7/4 find the value of A |
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Answer» SinA=60 or 120 1+$in square A=7/4=Sin squareA=7/4-1=Sin squareA=3/4=Sin A=√3/4=SinA=√3/2A=60 |
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| 33607. |
Can two numbers have 18 as their HCF and 380 as their LCM? Givw reason |
| Answer» LCM of 2 numbers is also a multiple of their HCF but in this case 380 is not a multiple of 18 so it is not possible to have 18 HCF and 380 as LCM of 2 numbers. | |
| 33608. |
_5+_8+_11 |
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| 33609. |
HelloSomebody give me latest year question papers |
| Answer» You can get them through this app☺ | |
| 33610. |
(K-12)x²+2(k-12)x+2=0 |
| Answer» {tex}D = [2(k\xa0-12)]^2 - 4(k\xa0- 12)\\times 2{/tex}= 4(k -12)2 - 8(k - 12)For equal and real roots, D = 0{tex}\\Rightarrow{/tex}{tex}4(k\xa0- 12)^2 - 8(k - 12) = 0{/tex}{tex}\\Rightarrow{/tex}{tex}4(k - 12) (k - 12 - 2) = 0{/tex}{tex}\\Rightarrow{/tex}{tex}4(k - 12) (k\xa0- 14) = 0{/tex}{tex}\\Rightarrow{/tex}k\xa0=12 or k\xa0= 14{tex}\\because{/tex}a\xa0{tex}\\ne{/tex}\xa00\xa0{tex}\\Rightarrow{/tex}\xa0k\xa0{tex}\\ne{/tex}12;\xa0{tex}\\therefore{/tex} k = 14. | |
| 33611. |
Why k is written in the formula for making quadratic equation |
| Answer» if we know A does not 0 so if we take constant k | |
| 33612. |
How to copy of my last year answer sheet |
| Answer» By the help of question paper | |
| 33613. |
Sir mujhe 10th,12th maths k previous papers chahiye.. please send me |
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Answer» 2017,2018 k paper nhi h Thanks can you please suggest me how can i download pdf please Yha pe koi sir ni h all r students like u bt u want previous paper that u\'ll download pdf ya fir aap "10 year" lelo isme pichle ds saal k ques paper hotte h y aapke nearest book depot p mill jayegi...? |
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| 33614. |
2018-2019 syllabus |
| Answer» Issi app m h | |
| 33615. |
Find the square of any positive integer 3q or 3q+1 |
| Answer» Ab q ki jagh m le lo as remainder and let common as q like(3q)2 =9q2 | |
| 33616. |
Find median of first 8 prime number |
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Answer» (11+7) ÷2 18÷2 9 First 8 prime numbers are 2,3,5,7,11,13,17,19Write these numbers and apply the formula of finding the median |
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| 33617. |
If p is a rational and √q is an irrational ,then prove that (p+√q) is an irrational. |
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| 33618. |
how to make under root 2 and5 |
| Answer» what rubbish is this | |
| 33619. |
how to make under root 5 |
| Answer» Students, plz write meaning full questions...... | |
| 33620. |
Simplify (3√5-5√2) (4√5+3√2) |
| Answer» (12×5)+9root 10 - 20 root 10 - (15×2) . . 60+9root 10 - 20 root 10 - 30 . 30+9root 10 - 20 root 10. 30-11root 10 . | |
| 33621. |
Write the additive inverse of2/8 |
| Answer» The additive inverse of 2/8 is -2/8 | |
| 33622. |
why i am fail |
| Answer» ur study was not upto the mark | |
| 33623. |
What is sure event? |
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Answer» sure event is the one that contains the whole sample space. For example, in our experiment of throwing a dice and noting the result, the event . Is it\'s possible is 1 |
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| 33624. |
Solve for the following equations for x and y. 7x +5y=74. 7x+1+5y+1 = 218 |
| Answer» So easy.....Itna easy bhi nahi aata | |
| 33625. |
Prove that one of any three consecutive positive integers must be divisible by3 |
| Answer» Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.By Euclid’s division lemma, we havea = bq + r; 0 ≤ r < bFor a = n and b = 3, we haven = 3q + r ...(i)Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.Putting r = 0 in (i), we get{tex}n = 3q{/tex}∴ n is divisible by 3.{tex}n + 1 = 3q + 1{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 2{/tex}∴ n + 2 is not divisible by 3.Putting r = 1 in (i), we get{tex}n = 3q + 1{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 2{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}∴ n + 2 is divisible by 3.Putting r = 2 in (i), we get{tex}n = 3q + 2{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}∴ n + 1 is divisible by 3.{tex}n + 2 = 3q + 4{/tex}∴ n + 2 is not divisible by 3.Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3. | |
| 33626. |
How to check copy |
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Answer» By the help of pen.... Honestly |
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| 33627. |
In how much days the rechecking result is declared |
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| 33628. |
If one of the zeros of the quadratic polynomial (a-1)x2 +ax+1 is -3 then find the value of a |
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Answer» -1 -1 a=-1 |
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| 33629. |
How can I fill rechaking form for class 10 |
| Answer» Pta nhi yaar me bhi wohi puch rha hu | |
| 33630. |
Show that 4m and 4m+2 Cannot be co-prime. |
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| 33631. |
Mare math ma compartment ha to fr sa kya tayrii kri jo pass ho jya |
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| 33632. |
N.c.e.r.t question class 11 |
| Answer» Check NCERT solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 33633. |
Hindi mai Hindi mai Bareilly |
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| 33634. |
Cheak wether 6n can end with the digit 0 |
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Answer» Silly questions No it cannot end with digit zero because it is not the factor of 5. The prine factorisation of 6n will be 2× 3 it does not contain 5. So the uniqueness of fundamental theorem of arithmetic guarantees that there are prime factorisation of 6......... |
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| 33635. |
if a and b are relatively prime then what is its HCF? |
| Answer» HCF of a and b is 1. | |
| 33636. |
What do you understand by linear combination? |
| Answer» A linear combination is an expression constructed from a set of terms by multiplying each term by a constant and adding the results (e.g. a linear combination of x and y would be any expression of the form ax + by, where a and b are constants). | |
| 33637. |
3x-4y-5 whole square |
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Answer» This app is to share knowledge and help others ,not to show yr attitude Aditya Good Q but i dont want to tell u |
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| 33638. |
Find the HCF of 65 and 117 and express at in the from 65m+177n |
| Answer» Behave yourself Aditya | |
| 33639. |
Find the zeroes of polynomial 4√3x+5x-2√3 |
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Answer» How?? 2_/3 /4_/3+5 Not possible... There will be 4√3x^2 Zeroes are -2/√3 and √3/4 |
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| 33640. |
If a boy got 469 out of 500 . So find the. percentage |
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Answer» 93.8 percentage 93.8% 21 94% |
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| 33641. |
Which date avialble rechecking form class 10 |
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| 33642. |
What is the square root of 25 |
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Answer» 5 5 5 5 5 5 |
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| 33643. |
If x=5/3and x=-1/2 are the zeroes of the polynomial ax2-7x+b then find the values of a and b. |
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| 33644. |
How can i download the form for applying to verification of marks |
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| 33645. |
1+tan^2a=sec^2a |
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| 33646. |
Types ofquadratic equation |
| Answer» The equations which have 2 as the degree of polynimial... i.e. in the equation highest degree must be 2 then the equation can be known as quadratic.... Like... ax2+bx+c=0 | |
| 33647. |
Excercise 5.1 |
| Answer» 3 | |
| 33648. |
Sin 60° cos 30°+cos 60°sin 30° evaluate |
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Answer» Sin 60° cos 30°+cos 60° sin 30°=0sin 60° sin (90-60) +cos 60° cos (90-60)=0sin 60° sin 60°+cos 60° cos 60°=01+1=2. Sin 60° cos(90-30) + cos 60° sin(90-30). =). Sin60°cos60° + cos 60° sin60° =). Sin^260° + cos^260° =). 1Bcz, sin^2theeta +cos^2theeta =1 |
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| 33649. |
Hou to represent pairs of linear equations in two variables on the graph |
| Answer» From the equations find out the values of x and y, from both the equations by making a box and putting any value in x and then finding out the value of y. In this way find out three values of x and y from both the equations. Then plot the points on the graph. | |
| 33650. |
Using quadratic formula solve x²– 2ax+3x–6a= 0 |
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