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Elan Coil [88]
2 years ago
11

1. Describe the impromptu speech. What are its major features?

Mathematics
1 answer:
Fantom [35]2 years ago
4 0

Answer: Down Below

Step-by-step explanation:

1. An impromptu speech, by definition, is the one that a speaker delivers without any prior preparation on the topic. Impromptu, itself, means “doing something without preparation”. In declamation contests, a random topic is fired at the speaker on the spot, and the speaker gets just a few seconds to think over the topic.

2. Lack of Gestures. The most common problem with public speakers is the statue like position, which they assume on the stage.

Lack of Energy.

No Rapport with Audience.

Use of Notes.

No Eye Contact.

Disrespect for Fellow Speakers.

Rushing with Words

3. Florida Strawberry Festival

The Florida Strawberry Festival is an 11-day community event celebrating the strawberry harvest of Eastern Hillsborough County. Each year, over 500,000 visitors enjoy the festival's headline entertainment, youth livestock shows, rides, exhibits of commerce and, of course, its strawberry shortcake.

4. Done without being planned, organised, or rehearsed.

5. What makes impromptu speech different from an extemporaneous speech?

The difference is in the delivery method: the impromptu speech is generated instantly and delivered immediately; whereas, the extemporaneous speech is delivered using just a few notes.

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A boat sails on a bearing of 038°anf then 5km on a bearing of 067°.
I am Lyosha [343]

This question is not complete

Complete Question

A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point

Answer:

a)8.717km

b) 54.146°

Step-by-step explanation:

(a)how far is the boat from its starting point.

We solve this question using resultant vectors

= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)

Where

Rcos θ = x

Rsinθ = y

= (4cos38,4sin38) + (5cos67,5sin67)

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x = 5.1056

y = 7.065

Distance = √x² + y²

= √(5.1056²+ 7.065²)

= √75.98137636

= √8.7167296826

Approximately = 8.717 km

Therefore, the boat is 8.717km its starting point.

(b)calculate the bearing of the boat from its starting point.

The bearing of the boat is calculated using

tan θ = y/x

tan θ = 7.065/5.1056

θ = arc tan (7.065/5.1056)

= 54.145828196°

θ ≈ 54.146°

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