The distance of the spaceship in discuss as in the task content given can be evaluated as; 800miles.
<h3>What is the distance the spaceship travels in 4 minutes?</h3>
The distance travelled by the spaceship in discuss can be evaluated by means of the slope of the linear relationship as follows;
Hence it follows from ratios that by observation, the linear relationship has a slope of 200mi/min.
Consequently, we can evaluate the distance travelled after 4 minutes as;
Distance = 200 × 4 = 800mi.
Ultimately, the distance travelled per minute by the spaceship is; 800mi.
Remarks:
600 miles
520 miles
800 miles
1,080 miles
Read more on ratios;
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Answer:
(1 cm)cos3πt
Step-by-step explanation:
Since the piston starts at its maximal height and returns to its maximal height three times evert 2 seconds, it is modelled by a cosine functions, since a cosine function starts at its maximum point. So, its height h = Acos2πft
where A = amplitude of the oscillation and f = frequency of oscillation and t = time of propagation of oscillation.
Now, since the piston rises in such a way that it returns to the maximal height three times every two seconds, its frequency, f = number of oscillations/time taken for oscillation where number of oscillations = 3 and time taken for oscillations = 2 s
So, f = 3/2 s =1.5 /s = 1.5 Hz
Also, since the the piston moves between 3 cm and 5 cm, the distance between its maximum displacement(crest) of 5 cm and minimum displacement(trough) of 3 cm is H = 5 cm - 3 cm = 2 cm. So its amplitude, A = H/2 = 2 cm/2 = 1 cm
h = Acos2πft
= (1 cm)cos2π(1.5Hz)t
= (1 cm)cos3πt
See photos for solutions and steps :)
Answer:
B
Step-by-step explanation:
Since this right angled triangle has two angles of 45, so according to theorem sides opposite to equal angles are equal
Applying Pythagoras theorem
(4√2)^2=(x)^2+(x)^2
32=x^2+x^2
32=2x^2
x^2=32/2
x^2=16
Taking sq root on both sides we get
x=4
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