How many 9-digit palindromes are there with all the digits being even and each digit appearing no more than twice?

96

Step-by-step explanation:

You can think a 9-digit palindrome as 5 slots.

In each slot, you can put a digit.

_ _ _ _ _

The first slot, is the 9th digit and the 1st digit.

The second slot is the 8th digit and the 2nd digit

etc.

This is for the nature of a palindrome number.

Then, we have 5 selections.

Now, remember that for the first four selections each digit will appear two times, so we can choose each even number only once.

Then:

For the first selection, we have 4 options (2, 4, 6 and 8. The zero can not be chosen here because the 9th digit can not be a zero if this is a nine digit number).

For the second selection we have 4 options, the 3 that remain for the previous selection and the 0, that is even.

for the third selection, we have 3 options.

for the second selection, we have 2 options

for the fifth selection we have only one option.

The total number of combinations is equal to the product of the number of options in each selection, this is:

C = 4*4*3*2*1 = 96

it shoild be o

step-by-step explanation:

x^2 + 81

we can factor this using difference of square binomials.

rewrite the equation:

x^2 - (-81)

factor:

(a^2 - b^2) = (√a^2 + √b^2)(√a^2 - √b^2)

(x^2 - (-81) = (x + √-81)(x - √-81)

we can remove the - from under the radical by multiplying it by i.

(x^2 - (-81) = (x + i√81)(x - i√81)

(x^2 - (-81) = (x + 9i)(x - 9i)

we can check our answer by using foil:

x^2 - 9ix + 9ix - 81i^2

x^2 - 81^2

i^2 = -1

x^2 + 81