Fully simplify only using positive exponents

3x^3y^6/6xy^3

To fully simplify the expression 3x^3y^6/6xy^3 using only positive exponents, we can simplify each term separately.

First, let's simplify the x terms: x^3/x = x^(3-1) = x^2.

Next, let's simplify the y terms: y^6/y^3 = y^(6-3) = y^3.

Now, substitute these simplified terms back into the expression: (3/6) * (x^2) * (y^3) = 1/2 * x^2 * y^3.

Therefore, the fully simplified expression is 1/2 * x^2 * y^3.

Fully simplify only using positive exponents

9x^4y^5/3x^8y^3

To fully simplify the expression (9x^4y^5) / (3x^8y^3) using only positive exponents, we can simplify the numerical coefficients and then simplify the variables separately.

First, let's simplify the numerical coefficients: 9/3 = 3.

Next, let's simplify the x terms: x^4 / x^8 = x^(4-8) = x^-4. Remember, a negative exponent indicates the reciprocal of the base, so x^-4 = 1/x^4.

Then, let's simplify the y terms: y^5 / y^3 = y^(5-3) = y^2.

Now, substitute these simplified terms back into the expression: 3 * (1/x^4) * y^2 = 3y^2/x^4.

Therefore, the fully simplified expression is 3y^2/x^4.

Write an equivalent expression by distributing the "-" sign outside the parentheses:

-(8.7f + 2g - 2)

To distribute the negative sign outside the parentheses, we can multiply each term inside the parentheses by -1:

-(8.7f + 2g - 2) = -8.7f - 2g + 2

Enter the missing values in the area model to find 5(4y - 5)


5 ?
?
-25

According to the model above, 5(4y - 5) =

According to the area model, we have:

5 (4y - 5) = 5 * 4y - 5 * 5

Simplifying, we get:

5 (4y - 5) = 20y - 25

What do you put in the model?

To complete the area model for 5(4y - 5), we would fill in the missing values as follows:

4y
5
-1
-25

This represents the expression 5(4y - 5) as the total area of the rectangle formed by multiplying the length (4y) by the width (5), and then subtracting the subtracting the outer rectangle formed by multiplying the width (5) by the height (-1). This gives us a final area of -25.

What’s the number underneath 4y?