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Let us assume that 3√2 is rational

Thus it can be in the form of 3√2=a/b

where a and b are Co prime numbers√2=(1/3)(a/b)

Here √2 is Irrational; a/3b is rational

Thus our assumption is wrong

Hence 3√2 is irrational

let x=0.33333......10x=3.33333.....9x=3x=3/9=1/3

find the value of determinants

S.I. = P×R×T ÷ 100

principal money + s.i.= amountamount is giventhen it is easy to find out time

isosceles have two sides equal. in this triangle both median and altitude is same. there fore point of concurrency of median and altitude i.e. orthocenter and centroid are same.

1.9*9*9

2.36*36 if you want any another there is also there

a=√3r=3/√3=3√3/3=√3let nth term will 729at^n-1=729√3×√3^n-1=729(√3)^n-1+1=(√3)^12(√3)^n=(√3)^12n=12

Let the number be xyvalue of number = 10x+y10x+y = 8(x+y)10x+y =8x+8y2x-7y=0....................(1)10x+y-45 = 10y+x9x-9y=45x-y = 5.................(2)multiply by -2-2x+2y=-10add to (1)-5y=-10y=2x-y =5x=7Number = 72

75,150,200find the LCM=750

| a c a+c || a+b b a | =4abc| b b+c c |

Expanding,

a[bc-a(b+c)]-c[c(a+b)-ab] +(a+c)[(a+b)(b+c)-(b)(b)]

=a[bc-ab-ac]-c[ca+cb-ab]+(a+c)[ab+ac+b²+bc]

=abc-a²b-a²c-c²a-c²b+abc+a²b+a²c+ab²+abc+abc+ac²+b²c+b²c

=4abc

LHS = (sinA-cosA+1)/(sinA+cos-1) = (1 + sinA) (1 - sinA)/cos A(1 - sinA) = 1 - sin2A/cos A(1 - sinA) = cos2A/cos A(1 - sinA) = cosA/(1 - sinA) = 1/ (1/cosA - sinA/cosA) = 1/(secA - tanA) = RHS Hence Proved

cos 60° = 1/2

cos 30° = √3/2

sin 60° = √3/2

sin 30° = 1/2

According to question,

cos 60° × cos 30° + sin 60° × sin 30°

cos ( 60° - 30°) = cos(30°) = 1/2

1)-441/21-21/1=-212)-195/-13=15/1=15

5000+2000-700+1000-500 = 6800So, Kalaivani's bank balance was Rs. 6800.

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6800 is correct answer

6800 is right answer .

6800 is the correct answer of the following

6800 is the correct answer

6800 is right answer of this question. please like my answer

5000+2000-700+100s-500=6800 is the right answre

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(729)^(1/3)=(9^3)^(1/3)=9(729×10^(-6))^(1/3)=(9^3×10^(-6))^(1/3)=9/100=0.09

thanks

nice answer

Step 1:Since 12576 > 4052, apply the division lemma to 12576 and 4052, to get

12576 = 4052 × 3 + 420

Step 2:Since the remainder 420 ≠ 0, apply the division lemma to 4052 and 420, toget

4052 = 420 × 9 + 272

Step 3:Consider the new divisor 420 and the new remainder 272, and apply thedivision lemma to get

420 = 272 × 1 + 148

Consider the new divisor 272 and the new remainder 148, and apply the divisionlemma to get

272 = 148 × 1 + 124

Consider the new divisor 148 and the new remainder 124, and apply the divisionlemma to get

148 = 124 × 1 + 24

Consider the new divisor 124 and the new remainder 24, and apply the divisionlemma to get

124 = 24 × 5 + 4

Consider the new divisor 24 and the new remainder 4, and apply the divisionlemma to get

24 = 4 × 6 + 0

The remainder has now become zero, so procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4.

Also, 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) =HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052)

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thanx

(-5)^14 × (125)^-12 × (25)^13 × (1/5)^16=(5)^14 × (5)^-36 × (5)^26 × (5)^-16= (5)^(14 - 36 + 26 - 16)= (5)^8

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1) 12)natural number3)irrational number 4)irrational number

(cos^2 20 + cos^2 70) + cot 25/tan 65 + cot 5 cot 10 cot 60 cot 80 cot 85

= (sin^2 (90-20) + cos^2 70) + tan(90-25)/tan 65 + tan(90-5) tan (90-10) cot 60 cot 80 cot 85

= (sin^2 70 + cos^2 70) + tan 65/tan 65 + tan 85 tan 80 cot 60 1/tan 80 1/tan 85

= 1 + 1 + cot 60

= 2 + 1/sqrt(3)

As we know digonal of parallelogram bisect each otherx-y = 5 .....(1)2y+4 = 8 y = 4/2 = 2so putting y = 2we get x =7

first tell the concept of this chapter

Ok. Parallelogram are type of quadrilateral in which opposite side are equal and parallel you can prove that digonal bisect each other thereby solve this problem.

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please question repeat please

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I)-15/30 = -5/10=1/-2;. ii) 16/-36=4/-9=2/-3. iii) -3/-15=1/5. iv) 68/119. = -1.5

(1) - 3/2(2) - 4/9(3) 1/5

324

5/2 answer is correct

This question is right answer 13.50

14/9 is the correct answer

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p(-1)=-2+3-8-2=1-10=-9

Let AB be the buildingAnd CD be the lamp post. While DE is the horizontal line parallel to the ground from the top of the lamp post to the building.So in Triangle ABC,AB=60mθ=60°tanθ= perpendicular /basetan 60°= AB / BC√3=60 / BCBC = 60 /√3On rationalising denominator, we getBC=60 * √3 /3 =[20 * √3]m

Distance between building and lamp post =20 √3 cm =20*1.732(√3=1.732) =34.64m(Ans)

EBCD is a rectangle , hence BC =EDIn Triangle AEDθ=30°tan 30° = AE / ED1 /√3 = AE / 20√3AE * √3 = 20√3AE = 20mAB = AE + EBEB =60 - 20=40mSince EB = CD (EBCD being a rectangle)CD or Height of lamp post = 40m (Ans)

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