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Dahasolnce [82]
3 years ago
7

How do you find the average speed for a non constant speeds? 

Mathematics
2 answers:
babunello [35]3 years ago
6 0
Average speed is calculated by dividing the total distance travelled by the time interval. For example, someone who takes 40 minutes to drive 20 miles north and then 20 miles south (to end up at the same place), has an average speed of 40 miles divided by 40 minutes, or 1 mile per minute (60 mph).
aleksandrvk [35]3 years ago
3 0
The average speed is the distance travelled divide by the time taken: Average speed = Distance travelled time taken
Hoped this helped
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Can someone please explain how to do this problem for me?
ololo11 [35]

Answer:

We always start with the x because in the alaphbet x is before y. So it says x=1 and the graph is skipping by 2 so in the middle of where the interact and beside positive 2 on the x row is where the 1 is so you keep it there for now. Now for the y you can see the middle where the interact well that is 0 so you have to keep the dot on the x place because if it said y=2 we would move it but is says 0 so we dont move it!

Pls Mark me brainlist

7 0
3 years ago
PLZ HELP!!! I Will give brainliest. What is the value of x in sin(3x)=cos(6x) if x is in the interval of 0≤x≤π/2
sertanlavr [38]

Answer:

sin(2x)=cos(π2−2x)

So:

cos(π2−2x)=cos(3x)

Now we know that cos(x)=cos(±x) because cosine is an even function. So we see that

(π2−2x)=±3x

i)

π2=5x

x=π10

ii)

π2=−x

x=−π2

Similarly, sin(2x)=sin(2x−2π)=cos(π2−2x−2π)

So we see that

(π2−2x−2π)=±3x

iii)

π2−2π=5x

x=−310π

iv)

π2−2π=−x

x=2π−π2=32π

Finally, we note that the solutions must repeat every 2π because the original functions each repeat every 2π. (The sine function has period π so it has completed exactly two periods over an interval of length 2π. The cosine has period 23π so it has completed exactly three periods over an interval of length 2π. Hence, both functions repeat every 2π2π2π so every solution will repeat every 2π.)

So we get ∀n∈N

i) x=π10+2πn

ii) x=−π2+2πn

iii) x=−310π+2πn

(Note that solution (iv) is redundant since 32π+2πn=−π2+2π(n+1).)

So we conclude that there are really three solutions and then the periodic extensions of those three solutions.

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