If x2 and x3 are meant to represent x2 and x3, then x2 times x3 = x5 You find the product of exponent variables by adding the exponents.
1.6667
x5 = x3 times x2. In this case x3 = 64 so x = cube root of 64 ie 4
x2(x3 + 1) is the best you can do there.
Andre Dawson x2, Bob Derier, Ron Santo x5, Greg Maddux x5, Mark Grace x2
x5 square units.
3²x5³x2²
x2 - x10 + x5 - x12 - x3
X5/X3 5 - 3 = X2 =====That is the exponent part; subtract the bottom exponent from the top exponent. X3/X5 3 - 5 X - 2 =====or, 1/X2 ===
(x1+x2+...+x5)/5 = 34 (x1+x2+...+x5)= 5 * 34 = 170 [(x1+x2+...+x5) + x6]/6 = 35 170 + x6 = 6*35 = 210 x6 =210-170=40 Answer is 40
A power can be factored in many different ways; for example: x5 = x times x4 = x2 times x3. When such a power appears as a term in combination with other expressions, you should look out for common factoring patterns. Here are two examples: x2 + x5 Here, you can use the pattern "common factor". In this case, both parts have the common factor x2, so you can factor it out. x4 - 1 Here, you can use the pattern "difference of squares". Note that x4 is the square of x2, and 1 is the square of 1.
-x(x2 - 2)(x2 + 3)