Q: Ed q2- q1 q1 q22 p2- p1 p1 p22 2 translates to what?

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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

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 âˆš25/âˆš9 = 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 x1âˆš3 and x2âˆš3. 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 =x1âˆš3)75 = (2x1)( x1âˆš3)/275 = x12âˆš3 (divide both sides by âˆš3)75/âˆš3 = x12A2 = b2h2/2 (replace A2 with 27, b2 = 2x2 and h2 =x2âˆš3)27 = (2x2)( x2âˆš3)/227 = x22âˆš3 (divide both sides by âˆš3)27âˆš3 = x22Since P1/P2 = x2/x2 also P12/P22 = x12/x22So that,P12/P22 = x12/x22P12/P22 = (75/âˆš3)/(27âˆš3)P12/P22 = 75/27 (square root of both sides)P1/P2 = âˆš75/âˆš27 = âˆš(75/27) = âˆš(25/9) = 5/3

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.

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.

p1+p2+p3=0

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ed=(q1-q2)/q1/(p1-p2)/p1

Ed=% Change in quantity demanded/% Change in price=(Q2-Q1)/Q1/(P2-P1)/P1= P1 - Price before change P2 - Price after change Q1 - Quantity before change Q2 - Quantity after change Ed- Price elasticity of demand

Ed=% Change in quantity demanded/% Change in price=(Q2-Q1)/Q1/(P2-P1)/P1= P1 - Price before change P2 - Price after change Q1 - Quantity before change Q2 - Quantity after change Ed- Price elasticity of demand

All mountain is only available in P1, but then there are the P bikes which are available in P1 and P2 The all mountain is more of an XC MTB, a bike you can go places with, while the P1 and P2 are more intended for bikeparks and similar use. They're about evenly spec-ed, but with the P1 single speed set up making it even more adapted for bikepark use.

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

P1 or parental

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In genetics, in a pure-breeding population, the parental generation is the P1 generation. The off-spring of the P1 Generation is called the F1 Generation

If the old population is P1, the new population is P2, and the growth rate is G, G = (P2 - P1) ÷ P1 x 100%

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#include#include#define n 100void del_char(char str1[n],char str2[]){char str[n],*p1,*q,i,j;p1=str1;q=str;*q=*p1;i=0;while(*p1!='\0'){{if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;if(*p1==str2[i]){p1++;i++;}else{q++;}}else{q++;}}else{p1++;q++;}*q=*p1;}i=0;}printf("%s\n",str);}main(){char str1[n],str2[]={"to "};clrscr();printf("please enter a text and includ 'to':\n");gets(str1);printf("remove 'to',remaining :\n");del_char(str1,str2);getch();return 0;}Output:please enter a text and includ 'the':Write a c program to simulate the calculator using functionremove 'the',remaining :Write a c program to simulate calculator using function

P1 Material is the name given to Carbon Manganese Steel in the engineering industry