3^2=9 9-4=5
How about: (3 squared minus 2 cubed) + 1 to the fourth 3 cubed minus 5 squared log 100 Kim Basinger's weeks minus Doris Day's cents.
32*5 - 6/2 = 9*5 - 6/2 = 45 - 3 = 42
After long division, (2x2-5x+5)/(x-2) = 2x-1+3/(x-2).
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(2-5)2 = (-3)2 = 9
3^2=9 9-4=5
A Huge ASS
It is impossible but if it were x squared plus 2x minus 15 the equation would be (x+5) (x-3) with x being equal to either -5 or 3. If the original problem was x squared minus 2x minus 15 the equation would be (x-5)(x+3) and x would be equal to either 5 or -3
How about: (3 squared minus 2 cubed) + 1 to the fourth 3 cubed minus 5 squared log 100 Kim Basinger's weeks minus Doris Day's cents.
32*5 - 6/2 = 9*5 - 6/2 = 45 - 3 = 42
This sequence is the result of 3 functions one after the other. These functions are: squared, add 2, minus 1. So 1 squared = 1, +2 = 3, -1 = 2, squared = 4, +2 = 6, -1 = 5, squared = 25.... So the next step would be +2 to get 27, then minus 1 to get 26.
It is 2n^3 + 7n^2 - 13n + 3
It is: 9-4 = 5
After long division, (2x2-5x+5)/(x-2) = 2x-1+3/(x-2).
-1ab^2 + 5b + 8 Step-by-step explanation: 3a^2+9ab+5-4a^2-4ab+3 3a^2-4ab^2=-1ab^2 9ab-4ab=5ab 5+3=8 -1ab^2+5ab+8
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