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Rufina [12.5K]
3 years ago
7

Someone Help me please??

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0
It will be identical, this will explain it https://www.wyzant.com/resources/answers/165599/how_many_degrees_can_a_regular_pentagon_be_rotated_a...
Tatiana [17]3 years ago
4 0
A 72° rotation would be 1/5 of the way around. So, turned counterclockwise, N would now be in position E.

There is your answer!
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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

4 0
3 years ago
Will Mark Brainlest helppp plss​
bearhunter [10]

Answer:

? hey your question is not full at all

7 0
3 years ago
Determine side x. Round to the nearest unit. 20 9 12 5
tatyana61 [14]

Answer:

12

Step-by-step explanation:

First, you will have to use trigonometry. Since the side of the triangleis the opposite of the angle 53 and the hypotenuse, you will use sin.

Sin 53 degrees = opposite side / hypotenuse side.

The opposite side is x and the hypotenuse side equals 15.

sin53= x/15

15 ( sin 53 ) =x

x= 11.97953

Round

x=12

6 0
3 years ago
Yo which set or sets does the Number 37 belong?
Tresset [83]
It is a whole number
3 0
3 years ago
How much icing is needed over the donut shown below (Diameter of 4in and 1.5in wide)
Dafna11 [192]
You use the formula C=2(3.14)r r is the radius and in this situation 2 so the answer will be 12.56
5 0
3 years ago
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