A body sliding on a smooth inclined plane requires 10 s to reach the bottom, starting from rest at the top. How much time does it take to cover initial one-fourth distance, starting with speed 2 m/s at the top? Length of incline is 8 m. The acceleration of a body sliding along incline depends on angle of incline and acceleration due to gravity only.

Answer:

Opposite of Worthwhile is Useless or Worthless.

Answer:

Explanation

worthless

useless

Answer:

Option (B)

Step-by-step explanation:

Who derived the formula for solving the quadratic equation ax²+bx+c = 0 ,where a≠0 ?

The standard quadratic equation is ax²+bx+c = 0

On dividing by a both sides then

=> (ax²+bx+c)/a = 0/a

=> (ax²/a)+(bx/a)+(c/a) = 0

=> x² +(bx/a) + (c/a) = 0

=> x²+(2/2)(bx/a) +(c/a) = 0

=> x²+(2bx/2a) +(c/a) = 0

=> x²+2(bx/2a) + (c/a) = 0

=> x² +2(b/2a)x = -c/a

=> x²+2x(b/2a) = -c/a

On adding (b/2a)² both sides then

=> x²+2x(b/2a)+(b/2a)² = (-c/a)+(b/2a)²

=> [x+(b/2a)]² = (-c/a)+(b²/4a²)

Since, (a+b)² = a²+2ab+b²

=> [x+(b/2a)]² = (-4ac+b²)/(4a²)

=> [x+(b/2a)]² = (b²-4ac)/(4a²)

=> x+(b/2a) = ±√[(b²-4ac)/(4a²)]

=> x+(b/2a) = ±[{√(b²-4ac)}/(2a)]

=> x = ±[{√(b²-4ac)}/(2a)] -(b/2a)

=> x = ±[{√(b²-4ac)}-b]/(2a)

=> x = [-b±√(b²-4ac)]/(2a)

Therefore, the roots are

[-b+√(b²-4ac)]/(2a) and [-b-√(b²-4ac)]/(2a).

This formula is known as Quadratic Formula .

This is derived by Indian Mathematician Sridharacharya.

Step-by-step explanation:

Answer:

Option (B)

Step-by-step explanation:

Corrected Question:-

Who derived the formula for solving the quadratic equation ax²+bx+c = 0 ,where a≠0 ?

Proof :-

The standard quadratic equation is ax²+bx+c = 0

On dividing by a both sides then

=> (ax²+bx+c)/a = 0/a

=> (ax²/a)+(bx/a)+(c/a) = 0

=> x² +(bx/a) + (c/a) = 0

=> x²+(2/2)(bx/a) +(c/a) = 0

=> x²+(2bx/2a) +(c/a) = 0

=> x²+2(bx/2a) + (c/a) = 0

=> x² +2(b/2a)x = -c/a

=> x²+2x(b/2a) = -c/a

On adding (b/2a)² both sides then

=> x²+2x(b/2a)+(b/2a)² = (-c/a)+(b/2a)²

=> [x+(b/2a)]² = (-c/a)+(b²/4a²)

Since, (a+b)² = a²+2ab+b²

=> [x+(b/2a)]² = (-4ac+b²)/(4a²)

=> [x+(b/2a)]² = (b²-4ac)/(4a²)

=> x+(b/2a) = ±√[(b²-4ac)/(4a²)]

=> x+(b/2a) = ±[{√(b²-4ac)}/(2a)]

=> x = ±[{√(b²-4ac)}/(2a)] -(b/2a)

=> x = ±[{√(b²-4ac)}-b]/(2a)

=> x = [-b±√(b²-4ac)]/(2a)

Therefore, the roots are

[-b+√(b²-4ac)]/(2a) and [-b-√(b²-4ac)]/(2a).

This formula is known as Quadratic Formula .

This is derived by Indian Mathematician Sridharacharya.

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Answer:

Explanation:

Young children are inherently capable of learning the necessary phonemes, morphemes, and syntax as they mature

When children start speaking they may not completely understand the nuances of a language. But I believe when these children start writing these languages they start understanding how particular these languages are.

By the age of 13 children can completely understand how a language is used. We know learning never stops. Even in the after ages one continuously learns about the language we use. But proper understanding can be developed once the child starts implementing the learning.

Answer:

1. take boiling tube and add copper sulphate crystals to it.
2. add some water to it.
3. heat it with the burner.

Answer:

The main use of crystallization in the organic chemistry laboratory is for purification of impure solids: either reagents that have degraded over time, or impure solid products from a chemical reaction

The apparatus required : filtering the mixture - conical flask, filter paper and filter funnel. heating the solution - water bath (beaker of water on a tripod and gauze, heated over a Bunsen burner) and evaporating basin. crystallisation - watch glass.

RESULT:-

Crystallisation process results in a physical change in objects. It leads to the formation of crystal structures. Clear distinguish between processes of crystallisation is not possible. But we can identify two categories of crystallisation processes namely Cooling Crystallisation and Evaporative Crystallisation.