h = 20t - 16t²

= (20 x 3/2) - 16 x (3/2)²

= 30 - ( 16 x 9/4)

= 30 - 36 = -6

The distance between the two points is 2√10 units .

According to the poet, the world will end due to the ‘fire’, which symbolises 'desire' but if the world has to end twice then it will be due to the hatred which is symbolised by ‘ice’. The poet feels that there is enough hatred spreading among the people in the world to perish the world twice because unlike fire, ice or hatred affects slowly and gradually and souls of people insensitive to anybody's pain.

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Both the statements are true but their meanings are different. Two people may have different developmental goals but they need not always be conflicting. For example, the developmental goal of different people may differ according to their status. A girl from a rich family may want the same freedom as her brother, may want to go on foreign trips, own a costly mobile, may want to study abroad etc. But a girl from a poor family will consider going to school and enjoying with friends as her developmental goal. Here the developmental goals are not conflicting.  An example of conflicting developmental goal can be the construction of a flyover to reduce waiting period at a railway crossing. This can be a developmental goal for an office goer. But if the construction of the flyover necessitates demolition of slums, it can be contradictory for the slum dwellers. But on the other hand, 24-hour electricity supply would be a developmental goal for almost each and every person.

फ्रैंकफर्ट जर्मनी की पहली, सभी के लिए, स्वतन्त्र रूप से चुनी गई, संसद थी। इसका चुनाव ०१ मई १८४८ को हुआ था। इसका सत्र १८ मई १८४८ से ३१ मई १८४८ तक फ्रैंकफर्ट एम मेन में सम्पन्न हुआ।

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The growth of industrialization changed the social and political equation of Europe in many ways, like:

1. Industrialization in its wake brought considerable change in socio-political sphere. Western and central parts of Europe witnessed rapid growth of industrialization leading to the growth of towns and cities and a strong commercial class whose existence was based on production for the market.
2. This industrialization also resulted in the emergence of working class population and a new middle class comprising industrialists, businessmen and professionals.
3. It was among these educated, liberal, middle classes that the ideas of national unity and abolition of aristocratic privileges gained popularity. All these changes had great impact on the society and political life of the people.

Let’s consider the number of 20 rupee notes to be x
And the number of 5 rupee notes be y

Then, according to the given conditions, we have

20x+5y=380     … (i) and

5x+20y=380−60

⇒5x+20y=320     … (ii)

Now, multiplying eq(i) by 4 and subtracting with eq(ii), we get

(80x+20y)−(5x+20y)=1520−320

⇒75x=1200

{tex}\begin{aligned} & \Rightarrow \mathrm{x}=\frac{1200}{75} \\ & \Rightarrow \mathrm{x}=16 \end{aligned}{/tex}

On substituting the value of x in eq(i), we get

20(16)+5y=380

⇒320+5y=380

⇒5y=380−320=60

{tex}\begin{aligned} & \Rightarrow y=\frac{60}{5} \\ & \Rightarrow y=12 \end{aligned}{/tex}

Therefore, 

The number of 20 rupee notes =16 and 

The number of 5 rupee notes =12

Given A circle with centre O and an external point T and two tangents TP and TQ to the circle, where P, Q are the points of contact.
To Prove: {tex}\angle{/tex}PTQ = 2{tex}\angle{/tex}OPQ
Proof: Let {tex}\angle{/tex}PTQ = {tex}\theta{/tex}
Since TP, TQ are tangents drawn from point T to the circle.
TP = TQ
{tex}\therefore{/tex} TPQ is an isoscles triangle
{tex}\therefore{/tex} {tex}\angle{/tex}TPQ = {tex}\angle{/tex}TQP = {tex}\frac12{/tex} (180^o - {tex}\theta{/tex}) = 90^o - {tex}\fracθ2{/tex}
Since, TP is a tangent to the circle at point of contact P
{tex}\therefore{/tex} {tex}\angle{/tex}OPT = 90^o
{tex}\therefore{/tex} {tex}\angle{/tex}OPQ = {tex}\angle{/tex}OPT - {tex}\angle{/tex}TPQ = 90^o - (90^o - {tex}\frac12{/tex} {tex}\theta{/tex}) = {tex}\fracθ2{/tex}= {tex}\frac12{/tex}{tex}\angle{/tex}PTQ
Thus, {tex}\angle{/tex}PTQ = 2{tex}\angle{/tex}OPQ

Solution is given below

सूरदास के पदों कि काव्यगत विशेषताएं :-

1) सूरदास के पदों में भक्ति भाव

का वर्णन है। सूरदास भक्तिकाल के कवि थे ।

तथा भक्ति उनके काव्य में आना जायस था ।

सूरदास भक्तिकाल के सगुण भक्ति के कृष्ण

भक्त शाखा के अन्तर्गत आते है ।वस्तुत: उनके

कव्यो में श्री कृष्ण के प्रति प्रेम का वर्णन

दिखाई देता है।

उदाहरण :-

हमारे प्रभु औगुन चित न धरौ।

समदरसी है मान तुम्हारौ, सोई पार करौ।

2) सूरदास के पदों में वात्सल्य चित्रण देखने

को भी मिलता है । सूरदास ने कृष्ण के बाल

अवस्था का वर्णन करने हेतु वात्सल्य रस का

ही प्रयोग किया है ।

उदाहरण :-

जसोदा हरि पालनैं झुलावै।

हलरावै दुलरावै मल्हावै जोइ सोइ कछु गावै॥

मेरे लाल को आउ निंदरिया काहें न आनि सुवावै।

तू काहै नहिं बेगहिं आवै तोकौं कान्ह बुलावै॥

3) सूरदास ने अपने पदों में अलौकिक प्रेम की

व्यजना किया है । उन्होंने गोपी - कृष्ण के

स्वरूप अलौकिक प्रेम को दर्शाया है ।

4) सूरदास के पदों में ब्रज भाषा का पुट

मिलता है। उन्होंने अपने काव्य में ब्रज भाषा

का प्रयोग किया है साथ ही ग्रामीण, अरबी -

फारसी जैसे शब्दो को भी अपने काव्य में

स्थान दिया है ।

5) सूरदास ने अपने काव्यो में ' पद ' छंद का

प्रयोग किया है। उन्होंने ' पद ' छंद को महत्व

दिया है ।

सूरदास के संबंध में विद्वानों का विचार :-

आचार्य हजारी प्रसाद द्विवेदी के अनुसार -

" सूरदास जब अपने प्रिय विषय का वर्णन

शुरू करते हैं तो मानो अलंकार-शास्त्र हाथ

जोड़कर उनके पीछे-पीछे दौड़ा करता है।

उपमाओं की बाढ़ आ जाती है, रूपकों की

वर्षा होने लगती है। "

With the help of dynamic mathematical models it is possible to forecast yields more accurately.

Verbal and non-verbal communication

Communication refers to the exchange of information, facts and feeling such that a common understanding is created.

Verbal communication in which communication takes place between the sender and the receiver by using words to transmit the message. It can be formal or non-formal type of communication. In this type of communication, presence of the sender and the receiver is not required.

Non-verbal communication in which communication takes place between the sender and the receiver by using signs. It can be proxemics, chronemics etc. In this type of communication, presence of the sender and the receiver is required.

To show that the perimeter of AABL is constant, we need to prove that AB + BA + AL + BL is constant.

First, let us label the angles in the figure as follows:

∠NLM = ∠LNM = α (since LM and LN are tangents from an external point, they are equal in measure)
∠MAB = β (angle between the tangent AB and the chord MC)
∠NBA = β (angle between the tangent AB and the chord NC)
∠ALN = γ (angle between the tangent LN and the chord MN)
∠BLM = γ (angle between the tangent LM and the chord MN)

Using the fact that the sum of angles in a triangle is 180 degrees, we can write:

∠LAM = 180 - α - β
∠LBN = 180 - α - β

Now, let us consider the perimeter of AABL:

AB + BA + AL + BL

Using the fact that angles in the same segment of a circle are equal, we can write:

∠MCN = ∠MAB + ∠NBA = 2β

Using the fact that the opposite angles in a cyclic quadrilateral add up to 180 degrees, we can write:

∠ALB = 180 - ∠MCN = 180 - 2β

Now, let us consider the triangles LAM and LBN. Using the fact that angles in a triangle add up to 180 degrees, we can write:

∠LAM + ∠ALN + ∠ALM = 180
∠LBN + ∠BLM + ∠BLN = 180

Substituting the values we have calculated for ∠LAM and ∠LBN, and using the fact that ∠ALN = ∠BLM = γ, we get:

(180 - α - β) + γ + ∠ALM = 180
(180 - α - β) + γ + ∠BLN = 180

Simplifying, we get:

∠ALM = α + β - γ
∠BLN = α + β - γ

Finally, let us consider the perimeter of AABL again:

AB + BA + AL + BL = AB + BA + 2AL sin γ + 2BL sin γ

Using the fact that AL = AM and BL = BN, and using the sine rule in triangles LAM and LBN, we get:

AB + BA + 2AM sin γ + 2BN sin γ = AB + BA + 2LM sin (α + β - γ) + 2LN sin (α + β - γ)

Using the fact that LM = LN (since they are tangents from an external point), we get:

AB + BA + 2LM sin (α + β - γ) + 2LM sin (α + β - γ)

Simplifying, we get:

AB + BA + 4LM sin (α + β - γ)

Using the fact that sin (α + β - γ) = sin (180 - γ) = sin γ, we get:

AB + BA + 4LM sin γ

Substituting the value of LM (which is equal to LN), we get:

AB + BA + 4LN sin γ

But we know that sin γ is a constant, since it is determined by the position of point L and the circle MN. Therefore, the perimeter of AABL is constant, since it does not depend on the position of point C on the arc MN.

Hence, we have proved that the perimeter of AABL is constant.

Trivial things can have great impact on our lives

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