How many 5 digit codes are possible if the first digit cannot be 0 or 1 and digits can be repeated?

80000

Step-by-step explanation:

The first digit is the only one with any constraints: it cannot be 1 or 0. This leaves us with 8 options (2, 3, 4, 5, 6, 7, 8, 9).

The other four digits have no such conditions, so we have 10 choices for each of them (0-9, inclusive).

So!

8 * 10 * 10 * 10 * 10 = 80000 :))

i think the answer is b)

step-by-step explanation:

rectangle: 9 1/8 x 4 1/9 = 37 37/72in

parallelogram: b times h= 10.3 x 6.2 = 63.86

trapezoid: a+b/2 times h= 42.5

volume of prism: l times w times h= 36