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zerowill be4t^2-5= 04t^2= 5t^2= 5/4t= +-√5/2

4t^2-5=0

We knowP( A u B ) = P(A)+P(B)-P(A n B)P( A n B' ) = P( A n B ) - P( B )

Given, P(A) = 2/3, P(B) =1/2, P(A n B) = 1/6

Then,P( A u B ) = 2/3 + 1/2 - 1/6 = (4 + 3 - 1)/6 = 6/6 = 1P( A n B') = 1/6 - 1/2 = (1 - 3)/6 = - 2/6 = - 1/3

12 multiplied by 12 equals to 144 144+24-10-18 equals to 140half of 140 is 70, then 70 bananas are in the basket

1 dozen is 12 bananas by multiply 12 =144(they are 12 square)added to 2dozens =24bananas10bananas get spoiltand after remov by 18there transmit from 2 bukets=144+

enter ans.=144+24=168168-10=158158-18=140total bananas is 140is equal to 2 bukets 140/2=70one bukets is 70andother bukets is 70

12 ×12 =144 ,144+12+12=168 then there is just said that bananas are spolit but not that they are removed . so we will subtract 18 only 168-18 =150 half of it is 75 and 75 .so the each basket have 75 bananas

12×12=144144+24=168168-10=158158-18=140140/2=70

nice solution

Common difference should remain constant in an AP.So, (6)-(2k+1) = (3k+1)-(6) 12 =(3k+1)+(2k+1) 12= 5k+2 10 =5k k=2.

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thank you so much...

Your answer is 18 bc you take 1 from the 2 add the one to the 5 so the 5 becomes a 15 then you do 15-7=8 then you bring down the 1 and add that to the 8 so it would be 18

The right answer is 18

the right answer is 18

y का मान ८० डिग्री होगा क्योंकि वह सम्मुख कोण हैतथा तिरभूज के तीनों कोणों का योग १८० डिग्री होता हैअतः x+80°+50°=180x= 180°-130°50°

5.

10%

As little as 10percentof theenergyat anytrophic levelistransferredto thenext level; the rest is lost largely through metabolic processes as heat.

300 is the right answer

700 and 300 is the correct answer of the given question

300 is the right answer of this question

700and 300is tha correct answer

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Suppose, S=1+2+3+............................. n (term)

Also S=n+(n-1)+(n-2)..........................3+2+1

adding that

2S=(n+1)+(n+1)+(n+1)+(n+1)+(n+1)+(n+1)+...........................(n+1)

2S =n.(n+1)

S=n.(n+1)/2

Hence proved.

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Plastic, the wonder material that we use for everything and which pollutes our environment, is perhaps the most harmful of trash dumped by mariners and sea-goers in sea.

Careless disposal of plastic can have dire consequences. A plastic bag looks like a tasty jellyfish to an indiscriminate feeder like the sea turtle, but plastic is indigestible. It can choke, block the intestines of, or cause infection in those animals that consume it.

A plastic bag can also clog an outboard engine’s cooling system. Lost or discarded monofilament fishing line can foul propellers, destroying oil seals and lower units of engines, or it can become an entangling web for fish, seabirds, and marine mammals.

According to the Centre for Marine Conservation, over 25,000 pieces of fishing line were collected from U.S. beaches during the 1996 annual beach clean-up and at least 40% of all animal entanglements reported during the clean-ups involved fishing line.

Every day, more and more plastic is accumulating in our oceans. Recreational boaters are not the only group that improperly disposes off plastic refuse at sea. Plastics also enter the marine environment from sewage outfalls, merchant shipping, commercial fishing operations, and beachgoers.

In the middle stage, it is very flexible and can be given any shape depending on temperature and pressure. In practices, urea, formaldehyde, poly ethylene, polystyrene, polycithylcholide, phenoloic compounds and other substances are used in the preparation of plastics pollution.

Polyvinyl chloride has also been found to destroy the fertility of the animals and their respiratory systems. When mixed with water, it causes paralysis and also damages bones and causes irritation to the skin.

Recently U.S.A. has banned the use of P.V.C. plastic in space apparatus and in food containers (as chemicals get dissolved in the food). India should immediately ban the use of P.V.C. in water pipes, food and medicine containers to save the lives of millions who are already suffering from different types of ailments.

LSA of cube is 4 × (side)² 4× (side)² = 576(side)² = 576/4 = 144side = √(144) = 12 cm

Total surface area is 6 × (side)²so 6× (12)² = 6 × 144 = 864 cm²

Volume of cube is (side)³ Volume = (12)³ = 1728 cm³

But its exactly different

Pls answer it again..

Solution :-

Diameter of the metallic cylinder = 5 cm

Radius = 5/2 = 2.5 cm

Height of the metallic cylinder = 3 1/3 cm = 10/3 cm

Volume of the metallic cylinder =πr²h

⇒ 22/7×2.5×2.5×10/3

⇒ 1375/21

Volume of the metallic cylinder = 65.476 cu cm

The metallic cylinder is melted and cast into a sphere. So, the volume of the metallic cylinder and the sphere will exactly same.

Volume of sphere = 4/3πr³

⇒ 65.476 = 4/3×22/7×r³

⇒ 88r³ = 65.476×21

⇒ 88r³ = 1374.996 or 1375 (Approx)

⇒ r³ = 1375/88

⇒ r³ = 15.625

⇒ r =∛15.625

⇒ r = 2.5

Hence, the radius of the sphere is 2.5 cm.

Then, diameter of the sphere = 2.5×2

= 5 cm

Volume of a cube is = (side)^3Thus we can write:(side)^3 = 5832 m^3taking cube roots both sides:side = 18m

not understood

a=3d=8-3= 5 tn=78tn=a+(n-1)d78=3+(n-1)575=(n-1)515=n-1n=16

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it's wrong answer is 1463:22

thanks buddy

by Taking √n comman

put in the limitsso as 1/infinity is 0 hence power will become zeroso anything raise to power zero is 1hence proved.

12 is the correct answer....

A bag contains = 97 kg of sugar so,4052 bag contain 4052×95=393044

the correct answer is 393044

your answer will be 9792

1:4 :: 9 :?

Let fourth proportion = x

We know,Product of means = Product of extremes

4 * 9 = 1* xx = 36

Therefore,Fourth proportion = 36

it is square os 1:2::3 hence last is square of 4 that is 16.hence 16 is answer.

x=9-1/4x=36-1/4x=35/4

Let theta = x

Given,cos x + sin x = √2cos x Then, (√2 -1)cos x = sin x

on multiplying both sides by (√2+1) , we get(√2+1)(√2-1)cos x = (√2+ 1) sin x⇒ cos x = √2sin x + sin x⇒ cos x - sin x = √2sin x

thank

In the cylinderof radius R, the sphere just fits. So its radius is R.

Volume of of cylinder up to the height 2R in the cylinder =πR² * 2R = 2π R³Volume of sphere = 4πR³/3

Volume of water in the gap between cylinder and sphere = 2πR³-4πR³/3 = 2πR³ /3

2) Depth of water in the can before sphere is put inside it = volume of water/area of cross section of can = (2π R³ / 3 )/π R² = 2 R/3 = 2 * 3.5 /3 = 7/3 cm

1) Surface area surface area of can in contact with water = flat surface + curved surface =πR² + 2π R * 2 R = 5π R² = 5 * 22/7 * 3.5² cm² = 110 /7 * (7/2)² cm² = 385 / 2 cm²