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88 + 8/8

88 + 8/8 = 89
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βˆ™ 2012-08-22 00:23:20
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: How can you use the digit 8 four times to make 89?
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Use the digit 4 four times to equal 6?

6 =(4+4)/4+4 problem solved :D

How many four digits numbers can be formed using 0123456789?

Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.

Use the digits 0 2 and 5 to make a three digit number that has both 2 and 5 as factors?


How many combinations can be made out of five digits?

Let's try and figure this one out...First case (the easy one) - We want to use any of the 10 digits to create a 5-digit number. In this case, we have 5 slots and in each slot we can put any of 10 digits.So the answer is 10*10*10*10*10 = 510 = 100,000 combinationsSecond case - We want to create a 5-digit number using any of the 10 digits, but we only want to use each digit once. This turns out to be only a little more difficult.Let's look at each digit individually.First digit - We can use any digit from [0-9], so we have 10 possibilities.Second digit - We can use any digit from [0-9] except for the one used in the first digit, so we now have 9 possibilities.Third digit - We can use any digit from [0-9] except for the ones used in the first and second digits, so we now have 8 possibilities.Fourth digit - As above, except we now have 7 possible choices.Fifth digit - As above, except we now have 6 possible choices.So our final number of combinations is 10 * 9 * 8 * 7 * 6 = 30,240 combinationsSpecial case - What happens if we want to create a number which does not begin with a zero? Well, we can make a simple adjustment to either of the above cases to take care of this. Just observe that the first digit is not limited to 9 possibilities [1-9], not 10.Special first case = 9 * 10 * 10 * 10 * 10 = 90,000 combinationsSpecial second case = 9 * 9 * 8 * 7 * 6 = 27,216 combinations

Use a 3 digit number to form all two digit numbers?

By using the 3 digits of a number we can form 3 different two digit numbers. 3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3

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