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How do you caluculate the GCD of two prime numbers?
Let p1 and p2 be the two prime numbers. Because they are prime, their divisors are div(p1) = {1,p1} and div(p2) = {1,p2}. So GCD(p1,p2) = Greatest Common Divisor of p1 and p2 = p1 if p1 equals p2 1 if p1 is different from p2
Can a multiple of a prime number be prime?
No. Let p1 be a prime number. Let p2 be a multiple of p1 such that p2 = p1 * k. Then the factors of p2 are: 1, p1, k and p2. ==> p2 is not a prime number. Hence, a multiple of a prime number cannot be a prime number.
What is the ratio of the perimiter of two equilateral tringles if their areas are 27 and 75?
We know that the perimeters of similar polygons have the same ratio as any two corresponding sides. Since we are dealing with equilateral triangles, we only need to find the scale factor because we know that in general, the ratio of the area of two similar figures is the square of the scale factor.Let A1 = 75 and A2 = 27. So we have,A1/A2 = 75/27 = 25/9Thus, the scale factor or the ratio of any two sides or the ratio of perimeters of the two triangles is √25/√9 = 5/3.Or find the side length of each equilateral triangle.In a triangle A = bh/2.Since we are dealing with equilateral triangles, we denote the bases of the triangles with 2x1 and 2x2, and the heights with x1√3 and x2√3. So that P1/P2 = 2x1/2x2 = x1/x2.Let's find these side lengths.A1 = b1h1/2 (replace A1 with 75, b1 = 2x1 and h1 =x1√3)75 = (2x1)( x1√3)/275 = x12√3 (divide both sides by √3)75/√3 = x12A2 = b2h2/2 (replace A2 with 27, b2 = 2x2 and h2 =x2√3)27 = (2x2)( x2√3)/227 = x22√3 (divide both sides by √3)27√3 = x22Since P1/P2 = x2/x2 also P12/P22 = x12/x22So that,P12/P22 = x12/x22P12/P22 = (75/√3)/(27√3)P12/P22 = 75/27 (square root of both sides)P1/P2 = √75/√27 = √(75/27) = √(25/9) = 5/3
When the ones and tens are multiplied separately each product is called a?
The general function is:1. y = a*x+bb is irrelevant and we can be removed2. y = a*xlets split x into ones and tens3. x = tens*10 + ones /e.g. 23 = 2*10 + 34. p1 = Multiplier of the onesp2 = Multiplier of the tens5. y = tens*10*p2 + ones*p1 /according to the question6. x*a = tens*10*p2 + ones*p1 /according to 2.7. (tens*10 + ones)*a = tens*10*p2 + ones*p1 /according to 3.8. tens*10*a + ones*a = tens*10*p2 + ones*p1 /regroup9. tens*10*a - tens*10*p2 + ones*a - ones*p1 = 0 /regroup10. tens*10*(a-p2) + ones*(a-p1) = 0 /regroup11. assuming "tens" and "ones" are not 0 then (a-p2) and (a-p1) must be 012. a-p2 = 0a-p1 = 013. a = p2a = p114. a = p1 = p2the answer is: when the Multipliers of ones and tens are equal then the product is called a.
What are the condition for three points to be collinear?
p1+p2+p3=0
