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11 months ago

A bolt has 6 1/2 turns per inch. How many turns would be in 2 1/2 inches of threads?

Mathematics

1 answer:

Reika [66]11 months ago

4 0

There would be 41 1/4 turns in 2 1/2 inches of threads.

Given, number of turns a bolt has = 16 1/2 turns per inch.

per inch bolt turns = 16 1/2 =33/2

how many turns would be there in 2 1/2 inches of threads = ?

Threads Per Inch, or TPI, is a measure of how many threads are found in one inch along a fastener's length. American fasteners are the only ones that employ TPI. Typically, the thread count is higher for smaller fasteners since they have finer threads. Just as the name implies, the Threads Per Inch (TPI) refers to the number of threads that run the length of a screw for one inch. The TPI of a screw can be easily calculated by simply counting the threads and dividing the total length.

so, 5/2 inches bolt turns =33/2 x 5/2

=165/4

hence 41 1/4 turns

Therefore, 2 1/2 inches of threads have 41 1/2 turns.

Learn more about Conversions here:

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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

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A football team has a probability of .75 of winning when playing any of the other four teams in its conference. If the games are

Alexeev081 [22]

Answer:

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Step-by-step explanation:

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Binomial probability distribution

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$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$