assignment
simplify. your answer should contain only positive exponents.
1) $2xy^2 \cdot 3y$
2) $4u^2v^3 \cdot 2uv$
3) $2ba^4 \cdot 2ba^4$
4) $(a^2b^3)^4$
5) $(4yx^3)^4$
6) $(3x^4y^4)^4$
7) $(4uv^3)^3$
8) $\frac{2a^3b^3}{3ab^4}$
9) $\frac{y^4}{3x^4y^3}$
10) $\frac{3u}{u^3v^2}$
11) $(4x^{-3}y^{-4})^{-4}$
12) $(2x^3y^4)^0$
13) $(x^3)^{-2}$
14) $(4a^3b^{-4})^{-4}$
15) $-\frac{x^{-3}y^3}{3x^4y^0}$
16) $\frac{2m^2n^0}{4mn^0}$
17) $-\frac{2u^{-1}v^{-3}}{2vu^3}$

simplify.
18) $\sqrt{180}$
19) $\sqrt{384}$
20) $\sqrt{27}$
21) $\sqrt{147}$
22) $\sqrt{24}$
23) $\sqrt{5} \cdot \sqrt{15}$

Question

assignment
simplify. your answer should contain only positive exponents.
1) $2xy^2 \cdot 3y$
2) $4u^2v^3 \cdot 2uv$
3) $2ba^4 \cdot 2ba^4$
4) $(a^2b^3)^4$
5) $(4yx^3)^4$
6) $(3x^4y^4)^4$
7) $(4uv^3)^3$
8) $\frac{2a^3b^3}{3ab^4}$
9) $\frac{y^4}{3x^4y^3}$
10) $\frac{3u}{u^3v^2}$
11) $(4x^{-3}y^{-4})^{-4}$
12) $(2x^3y^4)^0$
13) $(x^3)^{-2}$
14) $(4a^3b^{-4})^{-4}$
15) $-\frac{x^{-3}y^3}{3x^4y^0}$
16) $\frac{2m^2n^0}{4mn^0}$
17) $-\frac{2u^{-1}v^{-3}}{2vu^3}$

simplify.
18) $\sqrt{180}$
19) $\sqrt{384}$
20) $\sqrt{27}$
21) $\sqrt{147}$
22) $\sqrt{24}$
23) $\sqrt{5} \cdot \sqrt{15}$

Accepted Answer

Let's take problem 1: $2xy^{2}\cdot3y$ to solve.

# Explanation:
## Step 1: Multiply the coefficients
Multiply the numerical coefficients $2$ and $3$.
$2\times3 = 6$

## Step 2: Multiply the variables with the same base (using exponent rule $a^m \cdot a^n=a^{m + n}$)
For the $y$-terms: $y^{2}\cdot y=y^{2 + 1}=y^{3}$, and the $x$ term remains as it is.

## Step 3: Combine the results
Combine the coefficient and the variables.
$6\times x\times y^{3}=6xy^{3}$

# Answer:
$6xy^{3}$

Let's take problem 8: $\frac{2a^{3}b^{2}}{3ab^{4}}$ to solve.

# Explanation:
## Step 1: Divide the coefficients
Divide the numerical coefficients $2$ and $3$. The result is $\frac{2}{3}$

## Step 2: Divide the variables with the same base (using exponent rule $\frac{a^m}{a^n}=a^{m - n}$)
For the $a$-terms: $\frac{a^{3}}{a}=a^{3- 1}=a^{2}$
For the $b$-terms: $\frac{b^{2}}{b^{4}}=b^{2-4}=b^{-2}=\frac{1}{b^{2}}$ (but we want positive exponents, so we will handle this in the next step)

## Step 3: Combine the results and ensure positive exponents
Combine the coefficient and the variables. Since $b^{-2}=\frac{1}{b^{2}}$, we have $\frac{2}{3}\times a^{2}\times\frac{1}{b^{2}}=\frac{2a^{2}}{3b^{2}}$

# Answer:
$\frac{2a^{2}}{3b^{2}}$

Let's take problem 19: $\sqrt{384}$ to solve.

# Explanation:
## Step 1: Factor the number inside the square root into perfect squares and other factors
We know that $384 = 64\times6$ (since $64$ is a perfect square, $8\times8 = 64$)

## Step 2: Use the square - root property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ($a\geq0,b\geq0$)
$\sqrt{384}=\sqrt{64\times6}=\sqrt{64}\times\sqrt{6}$

## Step 3: Simplify the square root of the perfect square
Since $\sqrt{64} = 8$, we have $8\times\sqrt{6}=8\sqrt{6}$

# Answer:
$8\sqrt{6}$

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QUESTION IMAGE

Question

Explanation:

Step 1: Multiply the coefficients

Step 2: Multiply the variables with the same base (using exponent rule \(a^m \cdot a^n=a^{m + n}\))

Step 3: Combine the results

Step 1: Divide the coefficients

Step 2: Divide the variables with the same base (using exponent rule \(\frac{a^m}{a^n}=a^{m - n}\))

Step 3: Combine the results and ensure positive exponents

Step 1: Factor the number inside the square root into perfect squares and other factors

Step 2: Use the square - root property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\))

Step 3: Simplify the square root of the perfect square

Answer: