The idea is to add like terms together - in this case, the terms that have the variable "x" raised to the same power.
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.
4 (2x2) 9 (3x3) and 16 (4x4)
In a 4x4 square, you can fit a total of 16 1x1 squares, 9 2x2 squares, and 4 3x3 squares. This is calculated by considering the number of positions each square can occupy within the 4x4 grid. Specifically, a 1x1 square can occupy any of the 16 individual cells, a 2x2 square can fit into 9 different positions, and a 3x3 square can fit into 4 different positions.
Yes I can. I did it in QBasic about 15 years ago.
2x2=4 1x4=4 3x3=9 1x9=9
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
In a 4x4 square, you can fit a total of 16 1x1 squares, 9 2x2 squares, and 4 3x3 squares. This is calculated by considering the number of positions each square can occupy within the 4x4 grid. Specifically, a 1x1 square can occupy any of the 16 individual cells, a 2x2 square can fit into 9 different positions, and a 3x3 square can fit into 4 different positions.
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
Yes I can. I did it in QBasic about 15 years ago.