Question

Розв яжіть нерівніть:(2у-1)(3у+2)-6у(у-4)<48

Answer 1

1.    6y(кв)+ 4y-3y-2-6y(кв)+24у   2.   25у-2<48  и подставляй число.

Answer 2

I can help you with your math problem. To solve the inequality, you need to expand the brackets, collect the like terms, and find the critical points. Then you can use a sign chart to determine the intervals where the inequality is true. Here are the steps:

- Expand the brackets: `(2y-1)(3y+2)-6y(y-4) < 48` - Simplify the expression: `6y^2-15y-14 < 48` - Subtract 48 from both sides: `6y^2-15y-62 < 0` - Factor the quadratic: `6y^2-15y-62 = (3y+7)(2y-9)` - Find the critical points by setting each factor to zero: `3y+7 = 0` or `2y-9 = 0` - Solve for y: `y = -7/3` or `y = 9/2` - Make a sign chart with the critical points: `-7/3` and `9/2` - Test a value in each interval to see if the inequality is true or false: - For `y < -7/3`, use `y = -3`. Then `(3y+7)(2y-9) = (-2)(-15) = 30 > 0`. False. - For `-7/3 < y < 9/2`, use `y = 0`. Then `(3y+7)(2y-9) = (7)(-9) = -63 < 0`. True. - For `y > 9/2`, use `y = 5`. Then `(3y+7)(2y-9) = (22)(1) = 22 > 0`. False. - Write the solution in interval notation: `(-7/3, 9/2)`

I hope this helps you understand how to solve the inequality. If you have any other questions, feel free to ask me.Ответь+подробно.+Розв+яжіть+нерівніть:(2у-1)(3у+2)-6у(у-4)<48