Tell whether (-1,2) is a solution of 8x+y>-6

Arrange the steps to solve this system of linear equations in the correct sequence. x + y = -2 2x – 3y = -9 tiles subtract 3x + 3y = -6 (obtained in step 1) from 2x – 3y = -9 (given) to solve for x. substitute the value of x in the first equation (x + y = -2) to get y = 1. the solution for the system of equations is (-3, 1). x = -15 the solution for the system of equations is (-15, 13). add 3x + 3y = -6 (obtained in step 1) to 2x – 3y = -9 (given), and solve for x. x = -3 substitute the value of x in the first equation (x + y = -2) to get y = 13. multiply the first equation by 3: 3(x + y) = 3(-2) 3x + 3y = -6.

Nicolo paganini's numbers 1184 and 1210 are amicable. the proper divisors of 1184 are 1, 2, 4, 8, 16, 32, 37, 74, 148, 296, and 592. the proper divisors of 1210 are 1, 2, 5, 10, 11, 22, 55, 110, 121, 242, and 605. use the definition of amicable (friendly) numbers to show that they are indeed amicable. which of the folowing statement uses the definition of amicable numbers to show that 1184 and 1210 are indeed amicable? o a. o b. ° c. the sum of proper divisors of 1184 is 1210 and the sum of proper divisors of 1210 is 1184 the sum of proper divisors of 1184 is greater than 1210 and the sum of proper divisors of 1210 is greater than 1184 the sum of proper divisors of 1184 is less than 1210 and the sum of proper divisors of 1210 is less than 1184

Will mark brainliest, , and rate to only chrislaurencelleenzo is making a scale drawing of the rectangle below. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~imageenzo says that he can draw an enlarged rectangle that is 16 centimeters by 13 centimeters. which explains whether enzo is correct? enzo is correct because he used a factor of 2 to enlarge the rectangle.enzo is correct because he doubled one dimension and added the two lengths to get the other dimension.enzo is not correct because the enlarged rectangle should be 16 centimeters by 5 centimeters.enzo is not correct because he did not multiply the length and width by the same factor.