Anumerical pass code is required to open a car door. the pass code is five digits long and uses the digits 0–9. numbers may be repeated in the pass code. how many different pass codes exist

100,000

Step-by-step explanation:

A numerical pass code is required to open a car door.

The pass code is five digits long and uses the digits 0–9.

So, total number of possible digits will be 10

Condition:-

Total number of possible pass code = __×__×__×__×__

For each blank space we have 10 choices to fill because digit repeated. So, every blank space 10 possible digits.

Total number of possible pass code = 10×10×10×10×10

                                                            = 10⁵

                                                            = 100,000

Hence, Total number of pass code form will be 100,000

c.35

step-by-step explanation:

8+13=21

21+8=29

29+6=35