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Ans :- Option (b) is correct.

16+4-4+4×2=16 +4×2=20×2 =40

a=2 b=3(a+b)=(2+3)=5

polygon :is aplane figurethat is bounded by a finite chain of straight line segments closing in a loop to form a closedpolygon chainorcircuit. These segments are called itssides, and the points where two edges meet are the polygon'svertices orcorners.

Ans : (i) ,(ii)

Perimeteris the distance around a closed figure and is typically measured in millimetres (mm), centimetres (cm), metres (m) and kilometres (km). P=2πr

If one root of a quadratic equation (ax^2+bx+c) is the nth power of another, how do you prove that (a^n*c) + (a*c^n) +b=0?Let the two roots be α and α^n.

We know that sum of the roots is (-b/a) and product of the roots is (c/a).

(α)(α^n) = c/a

-> α^(n+1)= c/a

-> α = (c/a)^(1/(n+1))

α + α^n = -b/a

(Now let's replace α with the value we got)

(c/a)^(1/(n+1)) + (c/a)^(n/(n+1)) = -b/a

((a^n)c)^(1/(n+1)) + (a(c^n))^(1/(n+1)) = -b

((a^n)c)^(1/(n+1)) + (a(c^n))^(1/(n+1)) + b = 0

Therefore, value of ((a^n)c)^(1/(n+1)) + (a(c^n))^(1/(n+1)) = - b

other method

a/b+ b/a=a+b; a+b= (a)^2+(b)^2=a+b/ab;;

(a+b)^2/ab=a^2+ b^2; a^2+ b^2+2ab/ab=a^2+b^2=a^2+b^2

x तथा y ज्ञात कीजिए यदि [3

There is no feasible solution.

thanks

thanks oooooooooooooooooooooooooo

4a/(a²+1)≥1=> 4/(a²+1/a) ≥1=> 4/(a+1/a) ≥ 1=> 4 ≥ (a+1/a)

=> a+1/a ≤ 4

but also, a+1/a is always ≥ 2

so, only possible odd value between 2 and 4, is 3

oh i solved for value of a+1/a, here it is asking value of a

so just solve a+1/a = 3 => a²+1 = 3a=> a²-3a+1 = 0

here ∆ = (-3)²-4 = 9-4 =√5 so, 2 irrational real roots.. of a exist. => a = (3±√5)/2 or a = (3+√5)/2 and (3-√5)/2

2 solutions

edit - the process is perfectly correct.. do just assume that it is wrong by seeing the answer.

now I am unable to understand

Preference shares aresharesin theequityof a company that entitle the holder to a fixeddividendamount to be paid by theissuer. This dividend must be paid before the company can issue any dividends to its commonshareholders. Also, if the company is dissolved, the owners of preference shares are paid back before the holders ofcommon stock. However, the holders of preference shares do not usually have any voting control over the affairs of the company, as do the holders of common stock.

The types of preference shares are:

Callable. The issuing company has the right to buy back these shares at a certain price on a certain date. Since thecall optiontends to cap the maximum price to which a preference share can appreciate (before the company buys it back), it tends to restrict stock price appreciation.

Convertible. The owner of these preference shares has the option, but not the obligation, to convert the shares to a company's common stock at someconversion ratio. This is a valuable feature when themarket priceof the common stock increases substantially, since the owners of preference shares can realize substantial gains by converting their shares.

Cumulative. If a company does not have the financial resources to pay a dividend to the owners of its preference shares, then it still has the payment liability, and cannot pay dividends to its common shareholders for as long as that liability remains unpaid.

Non-cumulative. If a company does not pay a scheduled dividend, it does not have the obligation to pay the dividend at a later date. This clause is rarely used.

Participating. The issuing company must pay an increased dividend to the owners of preference shares if there is a participation clause in the share agreement. This clause states that a certain portion of earnings (or of the dividends issued to the owners of common stock) will be distributed to the owners of preferences shares in the form of dividends.

(4-d)³ + 4³ + (4+d)³ = 2884³-d³-12d(4-d) + 4³ + 4³+d³+12d(4+d) = 2883× 4³ + 12d² + 12d² = 28824d² = 288 - 19224d² = 96d² = 4d = √(4)d = 2 or - 2

option (3) is correct.

80%is the answer.

80% is the correct answer of the given question

80% is the correct answer

75 % is the correct answer

75

25/35=35/pp=(35×35)/25p=7×7=49

49 is right answer.............

(1) 5y-7=6yso -7=y(2) 4(2p-3)/3(4p-1)=7/38p-12=28p-77-12=20pso 20p=-5so p=-1/4

We know that

cos(2θ)=cos^2(θ)−sin^2(θ)cos(2θ)=1⋅(cos^2(θ)−sin^2(θ))cos(2θ)=(cos^2(θ)+sin^2(θ))(cos^2(θ)−sin^2(θ))[sincesin^2θ+cos^2θ=1]cos(2θ)=cos^4(θ)−sin^4(θ)[since(a+b)(a−b)=a^2−b^2]

Hence proved

thanks

x = a sin theta : y = b tan theta ....(Given)L.H.S. = a²/x² - b²/y² = a²/(a sin theta)² - b²/(b tan theta)²a² and b² are cancelled and then we get1/sin² theta - 1/sin² theta/cos² theta ...[∴ tan theta = sin theta/cos theta]1/sin² theta - cos² theta/sin² theta= (1-cos² theta)/sin² theta ....[∴ sin² theta + cos² theta = 1]sin² theta/sin² theta = 1= R.H.S⇒ a²/x² - b²/y²Hence proved.

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the increase is by 15 factor

15 is the right answer

15 is the right answer

I think 15 is the right solution of this question

I think 15 is the right solution of this question

16 F1=F2 is the correct answer for this question

the 15 is correct answer

15 is correct answer

16F1=F2 is the correct solution of this questions

40-(-39)+(-5)=40+39-5=40+34=74and40+(-39)-(-5)=40-39+5=40-34=6hence74>6

(1/30)÷(1/2)×75=(1/30)×2×75=5