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DedPeter [7]
3 years ago
14

The sequence 3, 5, 7 is a list of three prime numbers such that each pair of adjacent numbers in the list differ by two. Are the

re any more such "triplet primes"?
Mathematics
1 answer:
andreev551 [17]3 years ago
7 0

Answer:

  no

Step-by-step explanation:

Suppose the numbers of interest are x, x+2, x+4. If we divide x by 3, its remainder will be one of 0, 1, or 2.

If x mod 3 = 0, then x will be a multiple of 3, so won't be prime (unless x=3).

If x mod 3 = 1, then x+2 will be a multiple of 3, so won't be prime.

if x mod 3 = 2, then x+4 will be a multiple of 3, so won't be prime.

Hence the only three sequential odd numbers that are prime are 3, 5, and 7. There are no more such "triplet primes."

_____

Consequently, the name "prime triplet" is given to sequences of primes that have the largest and smallest differ by 6. The middle one differs from one of the other two by 2. It is conjectured there are an infinite number of that sort of prime triplet.

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The ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

Step-by-step explanation:

Let's call:

h the height of the ballonist above the ground,

a the distance between the two observers,

a_1 the horizontal distance between the first observer and the ballonist

a_2 the horizontal distance between the second observer and the ballonist

\alpha _1 and \alpha _2 the angles of elevation meassured by each observer

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S=a*h (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

S=S_1+S_2 (equation 2)

S_1=a_1*h

But we can write S_1 in terms of \alpha _1, like this:

\tan(\alpha _1)=\frac{h}{a_1} \\a_1=\frac{h}{\tan(\alpha _1)} \\S_1=\frac{h^{2} }{\tan(\alpha _1)}

And for S_2 will be the same:

S_2=\frac{h^{2} }{\tan(\alpha _2)}

Replacing in the equation 2:

S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})

And replacing in the equation 1:

h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}

So, we can replace all the known data in the last equation:

h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}\\h=\frac{7220 ft}{(\frac{1 }{\tan(35.6)}+\frac{1}{\tan(58.2)})}\\h=3579,91 ft

Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

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