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Alborosie
2 years ago
10

Find the remaining trigonometric functions of 0 if

Engineering
1 answer:
garik1379 [7]2 years ago
8 0

Answer:

cosΘ=−√558

tanΘ=−3√5555

cscΘ=83

secΘ=−8√5555

cotΘ=−√553

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Chemical engineering got is unofficial start around the time of the __________ __________ ________.
Gre4nikov [31]

Answer:

Option A,  World War II

Explanation:

During the period of industrial revolution around 1915-25, the chemical engineering has taken a new shape. During this period (i.e around the world war I), there was rise in demand for  liquid fuels, synthetic fertilizer, and other chemical products. This lead to development of chemistry centre in Germany . There was rise in use of synthetics fibres and polymers. World war II saw the growth of catalytic cracking, fluidized beds, synthetic rubber, pharmaceuticals production, oil & oil products, etc. and because of rising chemical demand, chemical engineering took a new shape during this period

Hence, option A is the right answer

4 0
2 years ago
It is required to design and implement: 1. A counter which counts from 0 to 255 with seven segment display.
liubo4ka [24]

Answer:

A counter which counts from 0 to 255 with seven segment display

Timer Mode Control (TMOD)

Explanation:

5 0
3 years ago
Digital information can be used to modulate an analog carrier wave. Like analog modulation, the carrier’s amplitude, frequency,
Fudgin [204]

Answer:

1) True, The carrier’s amplitude, frequency, and phase angle can be used to represent logical data (“1s” and “0s”).

Explanation:

6 0
3 years ago
Read 2 more answers
Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured
Paladinen [302]

Answer:  The normalized corrected value at 32.5 mm = 0.69 mV

Explanation:

Signal value V1 = 20.7 mV at

distance value = 29 mm and

V2 = 15.8 mV is measured at 32.5 mm. 

It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement

V1 = 20.7 - 5.1 = 15.6 mV

V2 = 15.8 - 5.1 = 10.7mV

The normalized corrected value at 32.5 mm will be

Vn = V2/V1 since V1 is the maximum value

Vn = 10.7/15.6 = 0.69 mV

3 0
3 years ago
Assume a 100-m tape measures 3 mm too short. Five measurements of a distance give results of 253.294, 253.287, 253.289, 253.284,
mariarad [96]

Answer:

attached below

Explanation:

5 0
2 years ago
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