1. How many 5-digit numbers are there for which the last digit is not a 6 or an 8, the digit 6 appears exactly once, and the number is even

8019 possible 5 digit numbers

Step-by-step explanation:

Available number to pick from are a total of 10 which are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

>> Now, let's consider when the first digit is 6 since we are told that 6 can appear only once.

For the first digit, it's only 1 way.

For the 2nd, 3rd, and 4th digit, we have 9 digits to pick from. Which is 9 ways each.

While for the 5th digit, it can't be 6 or 7. And it has to be even and so, we can pick only 0, 2, 4. Thus, we have 3 ways to choose.

So total number of ways here = 1 × 9 × 9 × 9 × 3 = 3 × 9³ ways

>> Now, let's consider when the 2nd digit is 6 since we are told that 6 can appear only once.

For the 1st digit, since 6 is not possible, and also 0 is not possible because a 5 digit number can't start with zero as it would no longer be a 5 digit number.

Thus, there are 8 ways to choose here.

For the second, third and fourth digits, there will be 9 ways each to chops.

While for the fifth digit, there will be 3 ways to choose.

Total number of ways here = 8 × 1 × 9 × 9 × 3 = 24 × 9² ways

>> When the third digit is 6, number of ways to choose will be same as above = 24 × 9² ways

>> When the fourth digit is 6, number of ways to choose will be same as above = 24 × 9² ways

We can't say none of the digits is 6 because we are told it appears exactly once.

Thus, total possible 5 digit numbers are;

(3 × 9³) + 3(24 × 9²) = 8019 possible 5 digit numbers

i you goofed up when writing this, because hnot is not a word minusf is not either. i cannot answer   because of that.

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!