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Yogita Ingle 6 years, 6 months ago
1 mole of ammonia weighs 17 g.
1 molecule of ammonia = (17 / 6.023 x 1023) = 2.82 x 10-23 g = 2.82 x 10-20 kg.
Posted by Rahul Dalal 6 years, 6 months ago
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Posted by Palak Agrawal 6 years, 6 months ago
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A-B=A∩Bc (A intersect B complement)
pick an element x
let x∈(A-B)
therefore x∈A but x∉B
x∉B means x∈Bc
x∈A and x∈Bc
x∈(A∩Bc)
x∈(A-B)
therefore A-B=A∩Bc
Posted by Piyush Bothra 6 years, 6 months ago
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Posted by Naveen Parnwal 6 years, 6 months ago
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Yogita Ingle 6 years, 6 months ago
(1) Two or more individuals or enterprises, either private or government, owned or a foreign company joining together through participation in equity capital for achieving a common target and mutual benefit is known as joint venture.
(2) Joint venture involves pooling of resources and expertise as well as sharing of risks and rewards by these enterprises.
(3) These enterprises agree to join together for the expansion of business, development of new products or for penetrating into new foreign markets.
(4) Joint ventures are formed either for long-term projects or for short-term projects.
(5) The basic purpose of joint ventures is to attain a strong position for both the enterprises.
Posted by Aman Sharma 6 years, 6 months ago
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Posted by Harshdeep Singh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
The integration of a function f(x) is given by F(x) and it is represented by:

Posted by Harpreet Singh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
tan(A - B) = [tan(A) - tan(B)]/[1 + tan(A)tan(B)].
Thus:
tan(13π/12)
= tan(4π/3 - π/4), from the hint
= [tan(4π/3) - tan(π/4)]/[1 + tan(4π/3)tan(π/4)], from the above formula
= (-1 + √3)/(1 + √3)
= -(1 - √3)/(1 + √3)
= -(1 - √3)^2/[(1 + √3)(1 - √3)], by rationalizing
= -(1 - √3)^2/(1 - 3), via difference of squares
= (1/2)(1 - √3)^2
= (1/2)(1 - 2√3 + 3)
= (1/2)(4 - 2√3)
= 2 - √3.
Posted by Prashant Kumar 6 years, 6 months ago
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Dhama ...? 6 years, 6 months ago
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Posted by Harpreet Singh 6 years, 6 months ago
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Posted by Kshitiz Chaudhary 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
since,sin60=√ 3/2
= √ 3/2( sin20sin40sin80)
=√ 3/2( sin20sin80sin40)
=√ 3/4 [(2sin20sin40)sin80]
on applying [cos(A-B)-cos(A+B) = 2sinAsinB]
we get,
= √ 3/4 (cos20-cos60)sin80 [since,cos(-a)=cosa]
= √ 3/4(cos20sin80-cos60sin80)
= √ 3/8(2sin80cos20-sin80)
= √ 3/8(sin100+sin60-sin80)
= √ 3/8( √ 3/2+sin100-sin80 )
= √ 3/8( √ 3/2+sin(180-80)-sin80 )
= √ 3/8( √ 3/2+sin80-sin80 ) [since,sin(180-a)=sina]
= √ 3/8( √ 3/2)
= 3/16
Posted by Harpreet Singh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
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Posted by Yash Goswami????? 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let P(n) = {tex}1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + .... + n(n + 1)(n + 2){/tex}{tex} = \frac{{n(n + 1)(n + 2)(n + 3)}}{4}{/tex}
For n = 1
{tex}P(1) = 1 \times 2 \times 3 = \frac{{1 \times 2 \times 3 \times 4}}{4} \Rightarrow 6 = 6{/tex}
{tex}\therefore {/tex} P (1) is true
Let P(n) be true for n = k.
{tex}\therefore P(k) = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ...{/tex}{tex} + (k + 1)(k + 2) = \frac{{k(k + 1)(k + 2)(k + 3)}}{4}{/tex} ... (i)
For n = k + 1
{tex}P(k + 1) = 1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + ... + {/tex}{tex}k(k + 1)(k + 2) + (k + 1)(k + 2)(k + 3){/tex}
{tex} = \frac{{k(k + 1)(k + 2)(k + 3)}}{4}{/tex} + (k + 1) (k + 2) (k + 3) [Using (i)]
{tex} = (k + 1)(k + 2)(k + 3)\left[ {\frac{{k + 4}}{4}} \right]{/tex}
{tex} = \frac{{(k + 1)(k + 2)(k + 3)(k + 4)}}{4}{/tex}
{tex}\therefore {/tex} P(k + 1) is true.
Thus P(k) is true {tex} \Rightarrow {/tex} P (k + 1) is true.
Hence by principle of mathematical induction, P(n) is true for all {tex}n \in N{/tex}.
Posted by Kartik Sharma 6 years, 6 months ago
- 3 answers
Dev Chauhan 6 years, 6 months ago

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Shiv Pratap Mishra 6 years, 6 months ago
1Thank You