Whats an even number with a 7 in the hundreds place, an odd number in the thousands place. and is a multiple of 10?

Our four-digit number is of the form ABCD

For example, if A = 2, B = 3, C = 5, D = 9, then ABCD = 2359. We are not multiplying the variables. The variables are simply digit placeholders.

A = thousands place value

B = hundreds place value

C = tens place value

D = ones or units place value

For a number to be even, the units digit must be 0,2,4,6, or 8. These are the possible values of D if we just wanted the number to be even. However, there's the further requirement that the four-digit number is also a multiple of 10. This means D must be 0. All multiples of 10 end with a zero.

So we know D = 0.

We're told that there is a "7 in the hundreds place", so B = 7.

We are also told that there is an odd number in the thousands place. We can pick from this list {1,3,5,7,9} for the value of A. I'll pick 1. So A = 1.

There is no restriction on what the tens place C must be, so pick any single-digit number you want for that. I'll pick C = 3.

We have: A = 1, B = 7, C = 3, D = 0

The number ABCD is therefore 1730

There are other possible answers because A and C can be multiple values. This is in contrast to B and D being fixed to one value only.

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step-by-step explanation:

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0