0
What else can I help you with?
What is (9x 12x3) (6x3 2x 4) simplified?
To simplify the expression ((9x \cdot 12x^3) \cdot (6x^3 \cdot 2x \cdot 4)), first calculate the coefficients: (9 \cdot 12 \cdot 6 \cdot 2 \cdot 4 = 5184). Next, combine the (x) terms: (x^1 \cdot x^3 \cdot x^3 \cdot x^1 = x^{1+3+3+1} = x^8). Thus, the simplified expression is (5184x^8).
Is (10x 36y) (2x y) 12x 36y?
To simplify the expression ((10x \cdot 36y) \cdot (2x \cdot y)), you multiply the coefficients and the variables separately. This results in (20x^2 \cdot 36y^2) which simplifies to (720x^2y^2). The expression (12x \cdot 36y) simplifies to (432xy). Therefore, ((10x \cdot 36y) \cdot (2x \cdot y)) is not equal to (12x \cdot 36y).
What is 3x 29 2x?
The expression "3x 29 2x" appears to be a multiplication of the terms involved. To evaluate it, you would multiply the coefficients and the variable parts: (3x \cdot 29 \cdot 2x = (3 \cdot 29 \cdot 2) \cdot (x \cdot x) = 174 \cdot x^2). Thus, the result is (174x^2).
How many numbers of permutations in the word embarrass?
The word "embarrass" has 9 letters, with the following frequency of letters: e (1), m (1), b (1), a (1), r (2), s (3). To find the number of permutations, we use the formula for permutations of multiset: [ \frac{n!}{n_1! \cdot n_2! \cdot n_3! \cdots} ] where ( n ) is the total number of letters and ( n_1, n_2, ) etc., are the frequencies of each letter. Thus, the number of permutations is: [ \frac{9!}{1! \cdot 1! \cdot 1! \cdot 1! \cdot 2! \cdot 3!} = \frac{362880}{1 \cdot 1 \cdot 1 \cdot 1 \cdot 2 \cdot 6} = 5040 ] Therefore, there are 5,040 distinct permutations of the letters in "embarrass."
What is 2(x 12)27x6- 120?
To simplify the expression (2(x \cdot 12)27x6 - 120), we first need to clarify any potential operations. Assuming (x) represents a variable and the multiplication is applied correctly, we can rewrite it as (2 \cdot 12 \cdot 27 \cdot 6 \cdot x^2 - 120). Calculating (2 \cdot 12 \cdot 27 \cdot 6) gives (432), so the expression simplifies to (432x^2 - 120).
What proprty is 5xp px5?
The expression (5xp \cdot px5) involves multiplying two terms. To simplify it, you can rearrange and group the terms: (5xp \cdot px5 = 5 \cdot p^2 \cdot x^2 \cdot 5). Therefore, it simplifies to (25p^2x^2).
How do you simplify 4x2x2 7x 3?
To simplify the expression (4x^2 \cdot 2 \cdot 7x \cdot 3), first, multiply the coefficients (4, 2, 7, and 3) together: (4 \cdot 2 \cdot 7 \cdot 3 = 168). Next, combine the (x) terms: (x^2 \cdot x = x^{2+1} = x^3). Therefore, the simplified expression is (168x^3).
What is 6x 3y 6z 3x 3z 5y?
To simplify the expression (6x \cdot 3y \cdot 6z \cdot 3x \cdot 3z \cdot 5y), first multiply the coefficients: (6 \times 3 \times 6 \times 3 \times 3 \times 5 = 4860). Next, combine the variables: (x^2) (from (x \cdot x)), (y^2) (from (y \cdot y)), and (z^2) (from (z \cdot z)). Thus, the simplified expression is (4860x^2y^2z^2).
What is 3x 2x in simplest form?
To simplify (3x \cdot 2x), you multiply the coefficients (3 and 2) and the variables (x and x). This results in (3 \cdot 2 \cdot x \cdot x = 6x^2). Therefore, the simplest form of (3x \cdot 2x) is (6x^2).
Which property is used for the following expression ( 78 ) 2 8 ( 2 7 )?
The expression ( (78) \cdot 2 \cdot 8 \cdot (27) ) appears to utilize the associative property of multiplication. This property allows for the grouping of numbers to be changed without affecting the product. For instance, rearranging or regrouping the factors, such as ( (78 \cdot 2) \cdot (8 \cdot 27) ), will yield the same result.
What happens to coefficients when you multiply variables?
When you multiply variables, the coefficients of those variables are also multiplied together. For example, if you have two terms, (a \cdot x) and (b \cdot y), and you multiply them, the resulting expression will be (a \cdot b \cdot (x \cdot y)). Thus, the coefficient of the resulting term is the product of the original coefficients.
What is another way to write 9x200?
Another way to write (9 \times 200) is (1800). You can also express it as (9 \cdot 200) or use the distributive property, such as (9 \cdot (2 \cdot 100)), which simplifies to (18 \cdot 100).
