PLEASE HELP How many 4-digit numbers divisible by 5, all of the digits of which are even, are there?

I assume you know Arithmetic Progression .

so, we have to find the first and last 4-digit number divisible by 5

first = 1000 ,  last =  9990

we have a formula,   = a + (n-1)d

here, is the last 4-digit number divisible by 5.

n is the number of 4-digit even numbers divisible by 5

d is the common difference between the numbers, which is 10 in this case

a is the first 4-digit number divisible by 5

9990 = 1000 + (n-1)*10

899 = n-1

n = 900

Hence, there are 900 4-digit even numbers divisible by 5

f-1(x)= cube root of 19x squared/x

step-by-step explanation: