Question

РЕШИТЕ ПОЖАЛУЙСТА)))) а)-20•(х-13)=-220; б)(30-7х)•8=352; в)5/12у-3/4=1/2; буду очень благодарна))),лавки вас^-^

Answer 1

А) - 20 * (х - 13) = - 220                 б) (30 - 7х) * 8 = 352
х - 13 = - 220 : (- 20)                      30 - 7х = 352 : 8 
х - 13 = 11                                       30 - 7х = 44 
х = 11 + 13                                      7х = 30 - 44
х = 24                                               7х = - 14
                                                          х = - 14 : 7
                                                          х = - 2
в) 5/12у - 3/4 = 1/2
5/12у = 1/2 + 3/4 = 2/4 + 3/4 = 5/4
у = 5/4 : 5/12 = 5/4 * 12/5 = 12/4
у = 3 

Answer 2

а) To solve the equation -20(x-13) = -220, we can follow these steps:

1. Distribute the -20 to the terms inside the parentheses: -20x + 260 = -220

2. Move the constant term to the other side of the equation by subtracting 260 from both sides: -20x = -220 - 260 -20x = -480

3. Divide both sides of the equation by -20 to isolate the variable x: x = -480 / -20 x = 24

Therefore, the solution to the equation -20(x-13) = -220 is x = 24.

б) To solve the equation (30-7x) * 8 = 352, we can follow these steps:

1. Simplify the equation by multiplying 8 to the terms inside the parentheses: 240 - 56x = 352

2. Move the constant term to the other side of the equation by subtracting 240 from both sides: -56x = 352 - 240 -56x = 112

3. Divide both sides of the equation by -56 to isolate the variable x: x = 112 / -56 x = -2

Therefore, the solution to the equation (30-7x) * 8 = 352 is x = -2.

в) To solve the equation 5/12y - 3/4 = 1/2, we can follow these steps:

1. Move the constant term to the other side of the equation by adding 3/4 to both sides: 5/12y = 1/2 + 3/4 5/12y = 2/4 + 3/4 5/12y = 5/4

2. Multiply both sides of the equation by the reciprocal of 5/12, which is 12/5, to isolate the variable y: y = (5/4) * (12/5) y = 60/20 y = 3

Therefore, the solution to the equation 5/12y - 3/4 = 1/2 is y = 3.

I hope this helps! Let me know if you have any further questions.