How many numbers can you form using the digits 1, 5, 6 and 9 without repetition of digits such that the value of the digit 6 in all those number is 100 times the value of the digit 6 in the number 9561?

6 times

Step-by-step explanation:

The value of 6 in the final number should be 100 times the value of 6 in 9561

in 9561 , the value of 6 is 60

100 times 60 is 6000

Hence, the value of 6 in the final number should be 6000

Now, we have the value of one digit out of 4 for our final number

So, our number will look like: 6 _ _ _

these dashes need to be filled by the remaining numbers: 1, 5 , 9

So, we will permute 1 , 5 and 9 in these 3 places since the position of 6 is permanent

3P3 = 3! = 6

Therefore, there are 6 possible 4-digit numbers which have the value of 6 as 6000 and no number are repeated from the given set of (1, 5, 6 , 9)

- 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

in the math equation 8 x 7 = 56, the 8 and the 7 are both factors of 56, since in this lesson we will use square tiles to figure out the factors of a whole find as many different arrays as you can, but follow one important rule: all the rows must have the i can make two rows of nine, so 2 and 9 are both factors of 18.

step-by-step explanation: