You would have to give four of them the 1/4 cookie and give the other to the fith one i you would have to be more specific please
Pascals triangle to the 6th:
1 x^0
1 1 x^1
1 2 1 x^2
1 3 3 1 x^3
1 4 6 4 1 x^4
1 5 10 10 5 1 x^5
1 6 15 20 15 6 1 x^6<span>
</span>the problem is to the 6th power so your going to use the 6th row of pascals triangle (don't count the first row). these numbers represent the coefficients of the variables
1(d-5y)^6 + 6(d-5y)^5 + 15(d-5y)^4 + 20(d-5y)^3 + 15(d-5y)^2 + 6(d-5y) + 1
then simplify
<h2><u>Answer: 9</u></h2><h2><u>Step-by-step explanation</u></h2><h2><u>Explanation:</u></h2><h2><u>Let n be the integer in question. Then we have</u></h2><h2><u>We now have a quadratic equation to solve. We could use the quadratic formula, however we know that </u></h2><h2><u>n</u></h2><h2><u> is an integer, so instead let's try to solve by factoring instead.</u></h2>
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Answer:
OMG this is so hard.OMG OMG OMG.But im 1% sure its 4
Step-by-step explanation:
Step-by-step explanation:
(x^3 - 9x) divided by (x^3 + x^2 - 6x)
= x(x^2 - 9) divided by x(x^2 + x - 6)
= x(x + 3)(x - 3) divided by x(x - 2)(x + 3)
= (x - 3)/(x - 2).
