Explore topic-wise InterviewSolutions in Current Affairs.

Suppose we consider, √6 is a rational number .

Then we can express it in the form of a/b

∴√6 = a/b, where a and b are positive integer and they are co-prime, i.e.HCF(a,b) = 1

∴√6 = a/b

=>b√6 = a

=>(b√6)² = a² [squaring both sides]

=>6b² = a²………..(1)

Here, a² is divided by 6

∴ a is also divided by 6. [we know that if p divides a², then p divides a]

∴ 6|a

=>a = 6c [c∈ℤ]

=>a² = (6c)²

=>6b² = 36c² [from (1)]

=>b² = 6c²

Here, b² is divided by 6,

∴ b is also divided by 6.

∴ 6|a and 6|b

we observe that a and b have at least 6 as a common factor. But this contradicts that “a and b are co-prime.”

It means that our consideration of “√6 is a rational number” is not true.

Hence, √6 is an irrational number.

I would use the proof by contradiction method for this.

So let's assume that the square root of 6 is rational.

By definition, that means there are two integers a and b with no common divisors where:

a/b = square root of 6.

So let's multiply both sides by themselves:

(a/b)(a/b) = (square root of 6)(square root of 6)a2/b2= 6a2= 6b2

But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even. So a2must be even. But any odd number times itself is odd, so if a2is even, then a is even.

Since a is even, there is some integer c that is half of a, or in other words:

2c = a.

Now let's replace a with 2c:

a2= 6b2(2c)2= (2)(3)b22c2= 3b2

But now we can argue the same thing for b, because the LHS is even, so the RHS must be even and that means b is even.

Now this is the contradiction: if a is even and b is even, then they have a common divisor (2). Then our initial assumption must be false, so the square root of 6 cannot be rational.

The switch and case keywords evaluate expression and execute any statement associated with constant expression whose value matches the initial expression. If there is no match with a constant expression, the statement associated with the default keyword is executed.

96 rupees only for the treatment

Total money= 12*4+5*8= 48+4088 thanks

why did india opt for planning

Your correct answer isa³-5a²+9a-8

sin(A/2).sin(7A/2) + sin(3A/2)sin(11A/2)

=(1/2) [ 2sin(7A/2).sin(A/2) + 2sin(11A/2)sin(3A/2) ]

=(1/2) [{ cos( 7A/2 - A/2) - cos( 7A/2 + A/2) } + { cos( 11A/2 - 3A/2)- cos( 11A/2 + 3A/2) } ( using formula, product to sum)

=(1/2) [ ( cos3A - cos4A ) + (cos4A - cos7A) ]

=(1/2) [ cos3A - cos7A ]

=(1/2) 2 sin[(7A - 3A)/2 ] sin[(7A+3A)/2 ] ( using formula, sum to product)

= sin2A sin5A ( Proved)

∫√x .dx=(x^3/2)×(2/3)

AB is the Chord. Radii drawn from points A & B are making 90 degrees at the center of the circle.

Thus ∆AOB is a right angl triangle with height and base both equal to radius ‘r’. = 14 cm

Area of the small segment.(Green Area) = Total are covered by sector AoB – Area of the triangle AOB.

Area of the sector AOB = (ᶿ/360) ∏*r^2

Since ᶿ = 90

Area of the sector AOB = (90/360) 22*14*14/7 = 154 Square cm

Area of the Triangle AOB = height * base / 2 = 14 * 14 / 2 = 98 Square cm. Area of the small segment.(Green Area) = 154 – 98 = 56 Square cm.

Area of the big segment = Area of the circle - Area of the small segment.(Green Area)

= 22*14*14/7 – 56 = 316 – 56 = 260 Square cm.

least number of toffees he should buy= lcm of three number5=1×510=2×512=2×2×3lcm=2×2×3×5=60

So, height of tower is 63.64 m and width of river is 31.82m.

uper side is this is 1st activity answer and down side is 2nd activity answer

let the height of the tower be hthen tan 30°=h/30h=30/√3=10√3m

1-81a²using identity a²-b²= 1-(9a²) = (1-9a) (1+9a)

Like if you find it useful

200 may be the answer

c) is correct answer

a) is the correct answer

a) the correct answer

b) a²+b²= c² is correct

He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.

1 friend can be selected out of 6 in 6C1 = 6 ways

2 friends can be selected out of 6 in 6C2 = 15 ways

3 friends can be selected out of 6 in 6C3 = 20 ways

4 friends can be selected out of 6 in 6C4 = 15 ways

5 friends can be selected out of 6 in 6C5 = 6 ways

6 friends can be selected out of 6 in 6C6 = 1 ways

Therefore the required number of ways (combinations) = 6 + 15 + 20 + 15 + 6 + 1 = 63

Each student will get64/8= 8 toffee sthanks

200*10/100=20200-20=180180*15/100=27180-27=153153+7=160208-160=148148/200*100=74profit=74%

profit is 74% I agree that I copied

discount = 200×10%= 200×10/100= 201st time cp = 200-20=180again he got 15% discountdiscount= 180×15/100= 27cp = 180-27= 153carrying charge= 7cp = 153+7=160sp = 208profit= 208-160=48%profit = profit ×100/cp = 48×100/160 = 30%

Tan(26°)=cot(90-26) = cot(64°)so 3tan^2(26)=3cot^2(64)so 3tan^2(26)–3cosec^2(64) can be written as3cot^2(64)–3cosec^2(64)= 3[cot^2(64)–cosec^2(64)] = 3(-1) = -3the answer is -3

sin(180+ A)= -sinAtan(360-A)= -tanAsec(180-A)= -sec Acos(90+A)= sinAnow-sin^3A*(-tanA)(sec^2A)/sin^2A*1/sin^2A*sinA

= tan^3A

sinx/ 1 + cosx + 1+ cosx/ sinx = sinx( 1 + cosx)^2/sinx( 1+ cosx) = . = sinx^2 + 1+ cosx^2+2cosx_/ sinx( 1+ cosx) = = 1 + 1 + 2cosx/sinx(1+ cosx) = 2 (1+ cosx)/ sinx( 1 + cosx)= = 2/ sinx = 2cosecx

thank u

sorry ,I cant understand near by #area of designs = 6 ×?????

My answer is 9months

solution

Ohhh simple ratio bee nai nikalna aata hai

Let B invested money for x months.4500 * 12 : 3000 * x = 2/1540/30x = 2/1540 = 60xx = 9 months.

the answer is 9 month......

9 month is the answer

the correct answer is 9 month

my answer is 9 month

the answer is 9 months

Please hit the like button if this helped you

thanks

Natural numbers are the set of positive integers, that is, integers from 1 to ∞ excluding fractional n decimal part. They are whole numbers excluding zero. Natural Numbers are also called counting numbers. Zero is the only whole number which is not a natural number.

OK, I'm assuming anyone reading this will be familiar with the concept a mod b... If you're not, a mod b is the remainder you'd get dividing a by b. I'm going to use = for "congruent to" here...

What this boils down to proving is that:

49^n + 16n = 1 mod 64

Let's start with the easy stuff...

Power of 49 49^n mod 640 11 492 333 174 1

WELL! Now, THAT simplifies things a lot, doesn't it? The index of 49 mod 64 is merely 4...

Thus, for all n, let n = a mod 4. Then, we can say that

49^n = 49^a mod 64.

OK, now all we have to do is get this 16 n thing in there...

When you do that, you get:

49^n mod 64 49^ n + 16n mod 641 149 49 + 16 = 6533 33 + 32 = 6517 17 + 48 = 651 1 + 64 = 65

In all of these cases, subtract 1 and you get 64...

thnqqqqqqq

(i) 9/100=0.09(ii)14/100=0.14The answer is correct.

9/100

Final result :

9/100 = 0.09

(1)9percent(2)14pecent

sin^-1{sinx}=x; iff x is in proper limits

we have to check wether x is in its proper limit i.e. -π/2<=x<=π/2Now,let nπ<x<(n+1)π

let,x=(n+1)π-y {iff x is more near to (n+1)π}x=nπ+y {iff x is more near to nπ}

since difference b/w max and min value of x is π so range of y is given by: 0<y<π/2

So, (n+1)π-x = y; is clearly in its proper limits[ since : 0<y<π/2]

Now, 3π<12<4π

sin12=-sin(4π-12)=sin(12–4π) {sin(2nπ-x)=-sinx}

Therefore, sin^-1{(sin12)} = sin^-1{sin(12–4π)}=12–4π

[ sin^-1{sinx}=x ; iff x is in proper limits]

sin^-1(sin12)=Sin^-1sin12=12

10+....+30+32+34 = (2+4+...+34)-(2+4+6+8) = 17/2(4+32) - 20 = 17*18 - 20 = 306 - 20 = 286

a^7/10*b^1y^-8/5÷a^-1/8b^5/4y^-5a^7/10+1/8*b^1-5/4*y^-8/5+5=a^56+10/80*b^4-5/4*y^-8+25/5=a^66/10*b^-1/4*y^17/5a^33/5*b^-1/4*y^17/5

Let us assume that5-√3 is rational.

Let ,5 -√3 = r , where "r" is rational

5 - r =√3

Here,LHS is purely rational.But,on the other hand ,RHS is irrational.This leads to a contradiction.Hence,5-√3 is irrational