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32251.

If the y=-2x+3 and y=3x-2 then the value of (x,y)

Answer» If y=-2x+3 & y=3x-2, then the value of (x,y) is We are given two equation in x and y. First equation => y = -2x + 3 Second equation => y = 3x - 2 Solving both equation,-2x + 3 = 3x - 25x = 5x = 1\xa0Substituting x = 1 in first equation,y = -2(1) + 3= -2 + 3= 1 Hence, ( x, y ) = ( 1, 1 )
32252.

Find roots of an equation (factorization) -6x²-x+2

Answer» 6x²+x-2 => 6x²-3x+4x-2 => 3x(2x-1)+2(2x-1) => (3x+2)(2x-1). => X=-2/3 or 1/2
32253.

Correct Explaination of Cross multiplication method??plzz from. Linear equation in ...

Answer»
32254.

Prove that 1 upon √5-√2 is an irrational number.

Answer» Let 1/2-√5 is a rational number.A rational number can be written in the form of p/q where p,q are integers.1/2-√5=p/q1/2-p/q=√5√5=(q-2p)/2qIf p,q are integers then (q-2p)/2q is a rational number.Then,√5 is also a rational number.But this contradicts the fact that √5 is an irrational number.So,our supposition is false.Therefore,1/2-√5 is an irrational numberHence proved.
32255.

What is the next number in series 6, 15, 35, 77, 143,.......?(A) 171(B) 151(C) 191(D) 181

Answer» 181
3 * 2=63*5 = 155*7 = 3511*7 = 7711*13 = 14313*17 = 221
32256.

(cosec A +sin A )(cosecA-SinA)=cot²A+Cos²A

Answer»
32257.

Show that 12^n constant end with digit 0 for any natural number n

Answer» If a number ends with zero then it would be divisible by 5 that is it\'s prime factorization must be 2*5. But this is not possible with 12n because its prime factorization is 2*2*3.It cannot be end with zero because its prime factors don\'t contain pair of 2*5
32258.

Write 500002 in international form

Answer»
32259.

Find thw HCF of 52 and 117and express it in the form of 52x +117y

Answer» See it is very very very very easy. Use your brain . If further you are not able to do then massage us i will tell you. By....
32260.

Class 7th worksheet in PDF chapter 2

Answer» U should go to class 7th
32261.

Prove that 3 is irrational number

Answer» Your question is wrong please recheck
32262.

Chapter 1st exercise 1.4 question number 2 you please explain

Answer» From the question no.1 we have to find the decimal expansion of only those numbers which have a terminating decimal expansion For example13÷3125 is terminating , then we have to find its decimal expansionIt can be written as 13÷5^5 and then we will multiply 2^5 upside and down 13*2^5÷5^5*2^5 13*b32÷10^5416÷10^50.00416
32263.

Prove that 1/2+√3 is irrational number

Answer» Can you prove that √3 is irrational?If you can, sum of rational and irrational is irrational
No
32264.

If A, B and C are interior angles of a triangle ABC, then show that sin [ B + C / 2 ] = cos A/2.

Answer» Then proved that sin[B+C/2]=cosA/2
And sin(90°-A/2)= CosA/2
Dividing both side by 2 ||. B+C/2=90°-A/2
We know that,A+B+C=180° ||. B+C=180°-A
???
32265.

Explain polynomial

Answer» a polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. An example of a polynomial of a single indeterminate, x, is x² − 4x + 7.
32266.

Prove that square of an odd integer decreased by 1 is a multiple of 8

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32267.

If sin theta equal to p/q then sec theta + tan theta equal to what

Answer»
32268.

HCF ka fourmla

Answer» Product of 2 no. by LCM
Here a and b is given +\'ve integers and q& r are unique whole no.
Product of twono./ LCM
We can find HCF by using Euclid\'s Division Lemma. a=bq+r , where 0 is equal or < r < b
32269.

3-√7Prove that. --------. Is a irrational number. 2+√7

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32270.

Write the. HCF of the smallest composite number and the smallest prime number

Answer» Abcd
2
32271.

Exercise 7.2

Answer»
32272.

Find the zeroes of polinomial 2xsq-5x

Answer» X=0 or 2/5
32273.

x+3y=6 2x-3y=12Draw a garph also

Answer»
32274.

2x square + 13 - 7

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32275.

Check whether the following are quadratic equation. 1.(x+1)2=(x-3)

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32276.

If two zeros if the polynomial (x2-6x3-26x2+138x-35) are (2+_/3) and 2-_/3) ,find other zeroes.

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32277.

Euclid dovision llemma

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32278.

ex 1.1 question no.3and4and5

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32279.

Eculid division lemme

Answer» Euclid’s Division Lemma:A lemma is a proven statement used for proving another statement.Theorem 1: “Given positive integers a & b, there exist unique integers q & r satisfying a = b*q + r, 0 ≤ r < b”.E.g. let’s assume that you have 34 apples & you have box that can accommodate 10 apples, then you have put these Apple in 3 boxes & you will have 4 apples remaining. Also note that the remaining apples are less than box size that is 10.34 Apples = 10 Apples * 3 Box + 4 Apples\xa0If you compare this with Euclid’s Division Lemma\xa0a = b*q + r, 0 ≤ r < b,\xa0Then a = 34, b=10, q=3 & r =4, 0 ≤ 4 <10
32280.

Chapter 1 exercise 1.1 question /solutions

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32281.

Find the polynomial when sum of its zeroes is 1/5and product of its zeroes is-1

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32282.

What is the HCF of 184,296 &1124 By Using Euclid Division lemma??

Answer»
32283.

Prove root 5 is irrational ?

Answer» Krish you are love guru
32284.

Exercise 2.4 NCERT solution

Answer»
32285.

sin2A=2sinAcosA

Answer» Sin(A+A)=sinAcosA+sinAcosA=> 2sinAcosA
32286.

Prove that product of two consecutive integers is divisible by 2

Answer» n(n+1)=n 2 +1=((n+1)−1)(n+1)=(n+1) 2 −(n+1)So if (n+1) is an even no (n+1) 2 is even and diff of even is always even is (n+1).If odd then (n+1) 2 is odd and diff of odd is always even So n(n+1) is always even and divisible by 2.
I want my answer now only
32287.

Arsarwal 9th ka1st chapter

Answer»
32288.

Prove that ³√5 is an irrational number.

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32289.

3x +4y=10 and 2x - 2y=2

Answer» 3x+4y=10→3x=10-4y → x=(10-4y)/3.......eq.1....similarly we find value of y by putting value if x which we find............so, 2x-2y=2→2[(10-4y)/3]-2y = 2→(now multiply2 by bracket) ,( 20-8y)/3=2+26→→20-8y = 6+6y→→ 20-6= 6y+8y→→ 14=14y→→ 14/14=y→→ 1 = y............hence y=1 .....now put value of y in eq1 ....x=[10-4(1)]/3→→ x=(10-4)/3→→x=6/3..→→x=2....,.Hence x=2 and y= 1 answer
32290.

Draw the graph of 4x+3y=24 , 2x-y=2

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32291.

The area of circular path of uniform width (h) surroundings a circular region of radius (r) is_

Answer» πh(2r+h)
I dont know
32292.

Practice paper question number 1

Answer»
32293.

Write fundamental theorem of arthemetic

Answer» It states that every integer greater than 1[3] either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to (except for) the order of the factors.
32294.

Prove underoot 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.
Ncert example...
32295.

Whatiszeroofpolynmials

Answer» Zeros of a polynomial can be defined as the points where the polynomial becomes zero on the whole. A polynomial having value zero (0) is called zero polynomial.\xa0In general, If k\xa0is zero of the linear polynomial in one variable; P(x) = ax +b, thenP(k) = ak+b = 0k = -b/a
The value which at which the equation becomes zero
32296.

What is euclid Algorithum and division???

Answer» Consider we have two numbers 78 and 980 and we need to find the HCF of both of these numbers. To do this, we choose the largest integer first, i.e. 980 and then according to Euclid Division Lemma, a = bq + r where 0 ≤ r ≤ b;980 = 78 × 12 + 44Now, here a = 980, b = 78, q = 12 and r = 44.Now consider the divisor as 78 and the remainder 44 and apply the Euclid division method again, we get78 = 44 × 1 + 34Similarly, consider the divisor as 44 and the remainder 34 and apply the Euclid division method again, we get44 = 34 × 1 + 10Following the same procedure again,34 = 10 × 3 + 410=4×2+24=2×2+0As we see that the remainder has become zero, therefore, proceeding further is not possible and hence the HCF is the divisor b left in the last step which in this case is 2. We can say that the HCF of 980 and 78 is 2.
32297.

X3- 8 (x-2)

Answer»
32298.

State euclids dividion lemma.

Answer» Given positive integers a and b,there exist unique integers q and r satifying a=bq+r, 0_< r
32299.

What is the name the founder

Answer» Please ask question with complete information.
32300.

35+35+35-400

Answer» _295
295
295