Question

(7z - 7/1)² - (24,5z + 11) (2z-1)>0​

Answer 1

$49z^2 - 98z + 49 - 49z^2 + 2,5z + 11 > 0 \\-95,5z + 60 > 0 \\191z z< \frac{60}{191}$

Answer 2

To solve the inequality $(7z - 7)^2 - (24.5z + 11)(2z - 1) > 0$, you can follow these steps:

1. Expand and simplify the expression on the left side of the inequality: $(7z - 7)^2 - (24.5z + 11)(2z - 1) > 0$ $49z^2 - 98z + 49 - (24.5z + 11)(2z - 1) > 0$

2. Continue to simplify: $49z^2 - 98z + 49 - (49z^2 - 24.5z - 22z + 11) > 0$

3. Distribute the negative sign and combine like terms: $49z^2 - 98z + 49 - 49z^2 + 24.5z + 22z - 11 > 0$

4. Simplify further: $49z^2 - 49z^2 - 98z + 24.5z + 22z + 49 - 11 > 0$

5. Combine like terms again: $-73.5z + 38 > 0$

6. Now, you want to isolate $z$, so first subtract 38 from both sides of the inequality: $-73.5z > -38$

7. Finally, divide both sides by -73.5, remembering to reverse the inequality sign since you're dividing by a negative number: $z < \frac{-38}{-73.5}$

Simplify the right side: $z < \frac{38}{73.5}$

Now you have the solution to the inequality: $z < \frac{38}{73.5}$

That's the solution for $z$.