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valentinak56 [21]
3 years ago
5

The table represents a linear function. The rate of change between the points (–5, 10) and (–4, 5) is –5. What is the rate of ch

ange between the points (–3, 0) and (–2, –5)?
Physics
2 answers:
Digiron [165]3 years ago
7 0
<span>8•5. Lesson 6. Lesson 6: Graphs of Linear Functions and Rate of Change ... in the table below: Input Output. 2. 5. 3.5. 8. 4. 9. 4.5. 10. ▫. The graph of a linear ... Yes, the rate of change betweeneach pair of inputs and outputs does seem to be .... The graph of the function will be a plot of fourpoints lying on a common line.</span><span>
</span>
NARA [144]3 years ago
3 0

Answer:

the anser is -5 hope you get a 100 on your test

Explanation:

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A beam of light bends when it passes from air into water, and then bends even more when it passes from water into glass. What ca
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Answer:

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4 0
3 years ago
If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t
Whitepunk [10]
From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)

Distance traveled is
d = 100 km = 10⁵ m

Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s

Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s

The difference in time is
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Answer:  283 s (nearest second)



6 0
3 years ago
Why does the spectroscopic parallax method only work for main sequence stars?.
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Answer:

Only main sequence stars have a well-defined relationship between spectral type and luminosity.

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5 0
2 years ago
A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an ini
Volgvan

Answer:

The horizontal range will be 2.55\times 10^5m

Explanation:

We have given initial speed of the shell u = 1.6\times 10^3m/sec

Angle of projection = 51°

Acceleration due to gravity g=9.8m/sec^2

We have to find maximum range

Horizontal range in projectile motion is given by

R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m

So the horizontal range will be 2.55\times 10^5m

6 0
3 years ago
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