Can you help me with my PreAlgebra??? I'm stuck on a problem?

Question

{(x2) (3x3) }3
all the numbers after the x are the small squared numbers

Answer 1

((x^2)(3x^3)^3=(x^2)^3(3)^3(x^3)^3
(x^6)(27)(x^9)=27x^(6+9)=27x^15

Answer 2

Usually, if we can't write the exponent above the number or variable it applies to, we use the ^ symbol to indicate an exponent. If so, your problem is written as {(x^2)(3x^3)}3.

Since order doesn't matter for multiplication and the parentheses and brackets are indicators of multiplication, we could rewrite this as x^2 * 3x^3 * 3 to make it a bit easier to see what's going on. When we multiply algebraic terms (the things separated by the multiplication symbol) we have to multiply all the coefficients (the "number parts") together, then combine the variables with the same bases. So our problem is 3 * 3 * x^2 * x^3. 3*3 is 9, x^2 * x^3 is x^5 (add the exponents) and putting the two together, you get 9x^5.

Answer 3

The way to write x squared here is x^2

I assume you want to simplify the expression

{(x^2)(3x^3)}^3

Inside the {} add the exponents for x

{3x^5}^3

Now (y^2)^3 = y^2y^2y^2 = y^6, so you multiply the exponents in this situation

{3x^5}^3 = 3^3x^15 = 27x^15

(The exponent of the 3 is an implied 1, and 1(3) = 3)

Answer 4

{(x2) (3x3)}3 =

{2X x 9}3 =

18X x 3 = 54x
x = 54

Answer 5

54

Answer 6

{x^2) (3 x (x^3)} 3

Answer 7

{(x2)(3x3)}3 = ( 2x (9) }3 = 2x (27) =54x

Answer 8

1st---> (x^2)(3x^3)= 3x^5
2nd---> (3x^5)(3)= 9x^5

sometimes it helps to replace the x with a number

Answer 9

Sorry.