Wiki User
∙ 10y agoThe idea is to add like terms together - in this case, the terms that have the variable "x" raised to the same power.
Wiki User
∙ 10y agoTo find the original number when given the prime factors, we have to work backwards. First, we simplify into 2 squared times 3 squared. (2x2) x (3x3) Then we expand the brackets. 4 x 9 = 54 Therefore 54 is the number, with 2x2 and 3x3 as the prime factors.
4 (2x2) 9 (3x3) and 16 (4x4)
2x2=4 1x4=4 3x3=9 1x9=9
1x1 = 1 2x2 = 4 3x3 = 9 4x4 = 16 ... Up to 31x31
Yes I can. I did it in QBasic about 15 years ago.
17 2x4=8+(3x3)=17 3x3=9 9+8=17
To find the original number when given the prime factors, we have to work backwards. First, we simplify into 2 squared times 3 squared. (2x2) x (3x3) Then we expand the brackets. 4 x 9 = 54 Therefore 54 is the number, with 2x2 and 3x3 as the prime factors.
Multiply the number by itself: 2x2=4, 3x3=9 and so on.
Multiply them. The number is 36.
No, it isn't. You can express 3x3-2x2 as 3x3-2x2+0x+0, so it actually has four terms. The definition of a binomial is an expression in the form Ax+b, where A and b are constants, so 3x3-2x2 is not a binomial. It is actually a quartomial.
72 ^ 9x8 ^ ^ 3x3 4x2 ^ 2x2
If you mean '2x4' as in 2 times 4- 2x4= 8 3x3= 9 4x2= 8 8-9-8=-19 if you want to divide by x -b/c x doesn't have a set amount- it would be: -19/x
4 (2x2) 9 (3x3) and 16 (4x4)
put 4 refined wood in the 2x2 grid in a square
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
i do not understand your question, do you mean: 2x2=4? 3x3=9? 4x4=16?
7x(3x3 - 2x2 + 5x + 1)