How many three-digit numbers greater than 200 can be formed from the digits 1, 2, 6, 7, and 9, if the digits can be repeated?if anyone can : ) 12 points

options:
a.) 64
b.) 100
c.)
125

100

B is correct

Step-by-step explanation:

If three-digit numbers greater than 200 can be formed from the digits 1, 2, 6, 7, and 9

Here we have total number of digits are 5

We need to make 3-digit number which is greater than 200

First digit must be 2 or greater than 2

From list of number 1,2,6,7,9

Four numbers are greater than equal to 2

For 1st place total number of possibility = 4

For 2nd place number of possibility = 5

For 3rd place number of possibility = 5

Total numbers of which are greater than 200 = 4 x 5 x 5

                                                                            = 100

Hence, 100 three-digit number greater than 200

- 2 + 10x³ - 21x² + 19x - 4

step-by-step explanation:

each term in the second factor is multiplied by each term in the first factor

x²(- 2x² + 4x - 1) - 3x(- 2x² + 4x - 1) + 4(- 2x² + 4x - 1) ← distribute

= - 2 + 4x³ - x² + 6x³ - 12x² + 3x -8x² + 16x - 4

collect like terms

= - 2 + 10x³ - 21x² + 19x - 4 ← in standard form

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step-by-step explanation:

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