Question

1. $-5x^{2}(3xy^{0}z)^{4}$

-405$x^{6}z^{4}$ blue
-15$x^{6}yz^{4}$ red
-405$x^{6}yz^{4}$ black
-405$x^{8}z^{4}$ silver

1. $-yx^{2}z^{3}(4y^{0}z)^{2}$

-4$x^{2}yz^{5}$ dark green
16$x^{2}yz^{5}$ orange
-16$x^{2}yz^{5}$ yellow
-16$x^{2}z^{5}$ green

1. $w^{-2}(4w)^{2}$

16$w^{4}$ yellow
16 gold
4$w^{2}$ orange
16$w^{2}$ light blue

1. $-3w^{-2} \cdot 2r^{-3}$

$\frac{-6}{w^{2}r^{3}}$ red
$\frac{-6}{w^{2}r^{3}}$ yellow
$\frac{-6w^{2}}{r^{3}}$ light pink
$\frac{-6r^{3}}{w^{2}}$ white

1. $\frac{4a^{2}bc^{3}}{24b^{-2}}$

$\frac{a^{2}c^{3}}{6b^{3}}$ pink
$\frac{6}{a^{2}b^{3}c^{3}}$ black
$\frac{a^{2}c^{3}}{6}$ light blue
$\frac{a^{2}b^{3}c^{3}}{6}$ orange

1. $\frac{(2p^{5}m)^{-2}}{p}$

4$m^{2}p^{11}$ gold
$\frac{m^{2}}{4p^{11}}$ orange
$\frac{1}{4m^{2}p^{11}}$ light blue
$\frac{4}{m^{2}p^{11}}$ purple

Explanation:

To solve the problem \(\frac{(2p^{5}m)^{-2}}{p}\), we follow these steps:

Step 1: Apply the power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). Applying this to the numerator \((2p^{5}m)^{-2}\):

$$ (2p^{5}m)^{-2} = 2^{-2}(p^{5})^{-2}m^{-2} $$

Step 2: Simplify the exponents

Using the power of a power rule \((a^m)^n = a^{mn}\), we simplify \((p^{5})^{-2}\):

$$ (p^{5})^{-2} = p^{5 \times (-2)} = p^{-10} $$

So now the numerator becomes:

$$ 2^{-2}p^{-10}m^{-2} $$

Step 3: Rewrite the negative exponents as positive exponents in the denominator

Recall that \(a^{-n} = \frac{1}{a^n}\). So we can rewrite \(2^{-2}\), \(p^{-10}\), and \(m^{-2}\) as:

$$ 2^{-2} = \frac{1}{2^2} = \frac{1}{4}, \quad p^{-10} = \frac{1}{p^{10}}, \quad m^{-2} = \frac{1}{m^{2}} $$

Putting it back into the numerator, we have:

$$ \frac{1}{4} \cdot \frac{1}{p^{10}} \cdot \frac{1}{m^{2}} = \frac{1}{4p^{10}m^{2}} $$

Step 4: Divide by \(p\) (which is the same as multiplying by \(\frac{1}{p}\))

Now we divide the numerator by \(p\):

$$ \frac{1}{4p^{10}m^{2}} \div p = \frac{1}{4p^{10}m^{2}} \cdot \frac{1}{p} $$

Step 5: Multiply the terms with the same base

Using the rule \(a^m \cdot a^n = a^{m + n}\) (in the denominator, since we have \(p^{10} \cdot p^1 = p^{10 + 1} = p^{11}\)):

$$ \frac{1}{4p^{10}m^{2}} \cdot \frac{1}{p} = \frac{1}{4p^{11}m^{2}} $$

Step 6: Compare with the given options

Looking at the options for problem 8, the expression \(\frac{1}{4p^{11}m^{2}}\) (or \(\frac{m^{-2}}{4p^{11}}\) which is equivalent) matches the LIGHT BLUE option \(\frac{1}{4m^{2}p^{11}}\) (since \(m^{-2} = \frac{1}{m^{2}}\)).

So the correct answer is the LIGHT BLUE option \(\frac{1}{4m^{2}p^{11}}\).

Answer:

To solve the problem \(\frac{(2p^{5}m)^{-2}}{p}\), we follow these steps:

Step 1: Apply the power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). Applying this to the numerator \((2p^{5}m)^{-2}\):

$$ (2p^{5}m)^{-2} = 2^{-2}(p^{5})^{-2}m^{-2} $$

Step 2: Simplify the exponents

Using the power of a power rule \((a^m)^n = a^{mn}\), we simplify \((p^{5})^{-2}\):

$$ (p^{5})^{-2} = p^{5 \times (-2)} = p^{-10} $$

So now the numerator becomes:

$$ 2^{-2}p^{-10}m^{-2} $$

Step 3: Rewrite the negative exponents as positive exponents in the denominator

Recall that \(a^{-n} = \frac{1}{a^n}\). So we can rewrite \(2^{-2}\), \(p^{-10}\), and \(m^{-2}\) as:

$$ 2^{-2} = \frac{1}{2^2} = \frac{1}{4}, \quad p^{-10} = \frac{1}{p^{10}}, \quad m^{-2} = \frac{1}{m^{2}} $$

Putting it back into the numerator, we have:

$$ \frac{1}{4} \cdot \frac{1}{p^{10}} \cdot \frac{1}{m^{2}} = \frac{1}{4p^{10}m^{2}} $$

Step 4: Divide by \(p\) (which is the same as multiplying by \(\frac{1}{p}\))

Now we divide the numerator by \(p\):

$$ \frac{1}{4p^{10}m^{2}} \div p = \frac{1}{4p^{10}m^{2}} \cdot \frac{1}{p} $$

Step 5: Multiply the terms with the same base

Using the rule \(a^m \cdot a^n = a^{m + n}\) (in the denominator, since we have \(p^{10} \cdot p^1 = p^{10 + 1} = p^{11}\)):

$$ \frac{1}{4p^{10}m^{2}} \cdot \frac{1}{p} = \frac{1}{4p^{11}m^{2}} $$

Step 6: Compare with the given options

Looking at the options for problem 8, the expression \(\frac{1}{4p^{11}m^{2}}\) (or \(\frac{m^{-2}}{4p^{11}}\) which is equivalent) matches the LIGHT BLUE option \(\frac{1}{4m^{2}p^{11}}\) (since \(m^{-2} = \frac{1}{m^{2}}\)).

So the correct answer is the LIGHT BLUE option \(\frac{1}{4m^{2}p^{11}}\).