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tankabanditka [31]
4 years ago
10

14n + p – 6n = 16p solve for and please explain

Mathematics
2 answers:
ValentinkaMS [17]4 years ago
7 0
The answer is in the bottom!!!
:)

mina [271]4 years ago
5 0

Answer:

14n+p-6n=16p

8n+p=16p

8n=15p

This doesn't really make sense,but this is what I got

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Find the perimeter of the figure
Sholpan [36]
The perimeter is the sum of all side lengths of a given shape, and can be computed by adding up all of the side lengths.

So in this case, it would be 15 x 2 + 13 x 2 the sum of them would be equal to the perimeter of the object, I believe.
3 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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