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3 years ago

Which of the following types of waves can move through water but cannot move through a vacuum where there is no matter?

Mathematics

2 answers:

Vikki [24]3 years ago

7 0

B is your answer!

Hope I helped!

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FLVS teacher,
~Amber Fish

solong [7]3 years ago

4 0

Let's consider each of the option....

Radio wave is an electromagnetic wave ; what is an electromagnetic wave?? An electromagnetic wave are waves that does not require material medium for their propagation... So radio wave can move through a vacuum because it's an electromagnetic radiation...

Sound wave is a mechanical wave ; what is a mechanical wave?? Waves that require material medium for their propagation.. Other examples include :water wave,

Light wave is also an electromagnetic wave(radiation) it does not require a material medium for its propagation it defines every other electromagnetic wave other example including... ; gamma ray, X-ray, ultraviolet ray, visible light, infrared Ray, radio wave(which I discussed first)

Sound wave can not move through a vacuum because it requires a material medium..
You can talk to someone and the person hears you because the air particles serves as the material medium....

Hope this helps..........

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Evgesh-ka [11]

Answer:

y = - 0.5 | x - 5 | + 2 

Step-by-step explanation:

g(x) = |x|

$1^{st}$ step - shifts right 5 units: g(x + 5) = | x - 5 |

$2^{nd}$ step - shifts up 2 units: g(x + 5) + 2 = | x - 5 | + 2

$3^{rd}$ step - reflected: - g(x + 5) + 2 = - | x - 5 | + 2

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y = - 0.5 | x - 5 | + 2

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3 years ago

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iris [78.8K]

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Alla [95]

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2 years ago

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Find the measure or indicated arc in P

ipn [44]

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Its 61

Step-by-step explanation:

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3 years ago

PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS

Reil [10]

Given

$x+1 = \sqrt{7x+15}$

We have to set the restraint

$x+1\geq 0 \iff x \geq -1$

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

$(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0$

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

$x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}$

So, we have to impose

$x-7\geq 0 \iff x \geq 7$

Squaring both sides, we have

$(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0$

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

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3 years ago