Question

(x+4)-7(x+4)+10; При x = одна вторая дробью; 0; -1.

Answer 1

Объяснение:

(х+4)-7(х+4)+10=(х+4)(1-7)+10=-6(х+4)+10=-6х-24+10=-6х-14

-6•1/2 -14=-3-14=-17

-6•0-14=-14

-6•(-1)-14=6-14=-8

Answer 2

To evaluate the expression (x+4)-7(x+4)+10 at specific values of x, you simply substitute the given values of x into the expression and perform the arithmetic calculations. Let's evaluate the expression for x = 1/2, x = 0, and x = -1.

1. When x = 1/2: (1/2 + 4) - 7(1/2 + 4) + 10 (9/2) - 7(9/2) + 10 (9/2) - (63/2) + 10 (9 - 63 + 20)/2 (-34 + 20)/2 -14/2 -7

2. When x = 0: (0 + 4) - 7(0 + 4) + 10 4 - 7(4) + 10 4 - 28 + 10 -14

3. When x = -1: (-1 + 4) - 7(-1 + 4) + 10 3 - 7(3) + 10 3 - 21 + 10 -8

So, the results of the expression (x+4)-7(x+4)+10 for the given values are as follows:

• When x = 1/2, the result is -7.
• When x = 0, the result is -14.
• When x = -1, the result is -8.