Explore topic-wise InterviewSolutions in Current Affairs.

ii) as 2 -0 is not equal to 4iv) as sqrt2 - 2× 4sqrt2 = - 7sqrt2 not equal to 4v) as 1-2 = -1 is not equal to 4

how 7 sqrt2

we get that because sqrt2 - 8sqrt2 = -7sqrt2

hi friend I don't know the answer of this question and please like my answer

I will solve your problem if you like this

1+secA. cosecA LHS =RHS

the series is.7,7+5, 12+5+2, 19+5+2+2, 28+5+2+2+2. therefore answer is 28.

2+2+2+5+28+2, 2+2+5+19, 2+5+12, 5+7+7 =25

the right answer is 28

the right answer of question is 28

28

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q where p and q are co- prime number.

So,√3 = p/q { where p and q are co-prime}√3q = p

Now, by squaring both the side

we get,(√3q)² = p²3q² = p² ........ ( i )

So,if 3 is the factor of p²then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sidesp² = (3m)²p² = 9m²

putting the value of p² in equation ( i )3q² = p² 3q² = 9m²q² = 3m²

So,if 3 is factor of q²then, 3 is also factor of q

Since, 3 is factor of p & q bothSo, our assumption that p & q are co-prime is wrong

hence, √3 is an irrational number

we can differentiate y wrt x and check its value is positive.

dy/dx= e^-x*x(2-x)nowdy/dx is positive when x(2-x) is positive as e^{-x} is always positive.

x (2 - x) is positive if 0 < x < 2 .

Hence y is increasing in the interval (0, 2)

Let us assume that√5 is a rational number.

We know that the rational numbers are in the form of p/q form where p, q are integers.so,√5 = p/q p =√5q

We know that 'p' is a rational number. So√5 q must be rational since it equals to pbut it doesnt occurs with√5 since its not an integer

Therefore, p =/=√5q

This contradicts the fact that√5 is an irrational numberhence our assumption is wrong and√5 is an irrational number.

Let us assume, to the contrary, that √5 is rational. i.e, we can find integer a and b such that √5= a/b.suppose a and b have a common factor other than 1,then we can divide by the common factor and assume that a and b are coprime. So, b√5=a.Squaring on both side, and rearranging, we get 5b^2=a^2.

Therefore , a^2 is divisible by 5, and by Theorem 1.3, it follows that a is also divisible by 3.So, we can write a =3c for some integer c. Substituting for a, we get 5b^2=25c^2, i.e , b^2 =3c^2.This means that b^2 is divisible by 5 and so b is also divisible by 5. Therefore, a and b have at least 5 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that √5 is rational. So, we conclude that √5 is irrational.

8(a-2b)(a-2b)-2(a-2b)-1suppose x=2a-b8x*x-2x-1=8x*x-4x+2x-1=4x(2x-1)+1(2x-1)=(4x+1)(2x-1)=(8a-4b+1)(2a-4b-1)

The answer of question is b)

the answer is B is right

The right answer is 'B'

Distance between pointsP(a+b, a-b) and Q(a-b, a+b)

D = sqrt[(a-b-a-b)^2 + (a+b-a+b)] = sqrt[(-2b)^2 + (2b)^2] = sqrt(4b^2 + 4b^2) = sqrt(8b)^2 = 2sqrt(2)b

Given:AD = CE

To prove:∠ABC = 1/2(∠DOE –∠AOC)

In△AOD and△COE

AD = CE (Given)

AO = OC and DO = OE (Radii of same circle)

∴△AOD ≅ △COE (By SSS congruence criterion)

⇒∠1 =∠3,∠2 =∠4 (CPCT) ....(i)

But OA = OD and OC = OE ⇒∠1 =∠2 and∠3 =∠4 .....(ii)

From (i) and (ii), we have

∠1 =∠2 =∠3 =∠4 (= x say)

Also, OA = OC and OD = DE

⇒∠7 =∠8 (= z say) and∠5 =∠6 (= y say)

Now, ADEC is a cyclic quadrilateral

⇒∠DAC +∠DEC =180°

⇒ x + z + x + y =180° ⇒ y =180°- 2x - z ...(iii)

In△DOE,∠DOE = 180° - 2y

and in△AOC,∠AOC = 180° - 2z

∴∠DOE -∠AOC = (180° - 2y) - (180° - 2z) = 2z - 2y

= 2z - 2(180° - 2x - z) (Using (iii))

= 4z + 4x - 360° ....(iv)

Again, ∠BAC + ∠CAD = 180° ⇒∠BAC = 180°– (z + x) ....(v)

Similarly, ∠BCA = 180°– (z + x) ....(vi)

In△ABC,∠ABC =180° -∠BAC - ∠BCA

=180° - 2[180° - (z - x)] (Using (v) and (vi))

= 2z + 2x -180° = 1/2(4z + 4x - 360°) ....(vii)

From (iv) and (vii), we have

∠BAC = 1/2(∠DOE - ∠AOC)

10-[15-{8-6-8-2}]=10-[15-{-8}]=10-[15+8]=10-23=-13

6+(8-2)=6+6=12(5+3)-(4*2)8-8=0

Thanks alot

Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

Now, by squaring both the side

we get,

(√3q)² = p²

3q² = p² ........ ( i )

So,

if 3 is the factor of p²

then, 3 is also a factor of p ..... ( ii )

=> Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

3q² = p²

3q² = 9m²

q² = 3m²

So,

if 3 is factor of q²

then, 3 is also factor of q

Since

3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

hence,. √3 is an irrational number

Hit the like 👍 ✔️👍

1

2

3

Centroid = (x1+x2+x3)/ 3, (y1+y2+y3)/3= (-7+2+8)/3 , (6-2+5)/3 = 3/3 , 9/3 = 1,3

So,centroid = (1,3)

centroid = (x1+x2+x3/3 , y1+y2+y3/3)

= -7+2+8/3 , 6-2+5/3

= 3/3 , 9/3

= 1,3