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$2^{-8}=\\frac{1}{2^8}=\\frac{1}{256}$; $a^7\\cdot b^{-3}\\cdot c^0=\\f…

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$2^{-8}=\frac{1}{2^8}=\frac{1}{256}$; $a^7\cdot b^{-3}\cdot c^0=\frac{a^7}{b^3}$; $16^{-2}=\frac{1}{16^2}=\frac{1}{256}$; $7b^4\cdot a^{-2}\cdot d^6=\frac{7d^6b^4}{a^2}$; $-1\cdot k^0\cdot d^{-3}\cdot 4d^6=4d^9$ (with steps $1 - (-1)=0$? wait, maybe exponent rules: $d^{-3}\cdot d^6 = d^{3}$, $-1\cdot4=-4$? wait original has $-1\cdot k^0\cdot d^{-3}\cdot 4d^6 = 4d^9$ and $d^6 - d^{-3}=d^9$? maybe typos, but ocr text: $2^{-8}=\frac{1}{2^8}=\frac{1}{256}$; $a^7\cdot b^{-3}\cdot c^0=\frac{a^7}{b^3}$; $16^{-2}=\frac{1}{16^2}=\frac{1}{256}$; $7b^4\cdot a^{-2}\cdot d^6=\frac{7d^6b^4}{a^2}$; $-1\cdot k^0\cdot d^{-3}\cdot 4d^6=4d^9$; $1 - (-1)=0$; $d^6 - d^{-3}=d^9$; $(5\cdot10^3)\cdot(0.2\cdot10^{-4})=\frac{1}{10^1}$; $\frac{5}{0.2}=1$; $\frac{10^3}{10^{-4}}=10^{-1}$; $(4\cdot10^0)\cdot(3\cdot10^{-2})=\frac{12}{10^2}$; $\frac{4}{3}=12$? wait no, $4\cdot3=12$, $10^0\cdot10^{-2}=10^{-2}$; $(2\cdot10^2)\div(7\cdot10^{-1})=\frac{3.5}{10^{-5}}$; $\frac{7}{2}=3.5$; $\frac{10^2}{10^{-7}}=10^{-5}$

Explanation:

Step1: Analyze the first expression \(2^{-8}\)

Recall the negative exponent rule: \(a^{-n}=\frac{1}{a^{n}}\). So \(2^{-8}=\frac{1}{2^{8}}\). Calculate \(2^{8}=256\), so \(2^{-8}=\frac{1}{256}\).

Step2: Analyze the second expression \(16^{-2}\)

Using the negative exponent rule, \(16^{-2}=\frac{1}{16^{2}}\). Calculate \(16^{2}=256\), so \(16^{-2}=\frac{1}{256}\).

Step3: Analyze the expression \(a^{7}\cdot b^{-3}\cdot c^{0}\)

Recall the zero - exponent rule \(a^{0}=1\) (\(a
eq0\)) and the negative exponent rule. So \(c^{0}=1\) and \(b^{-3}=\frac{1}{b^{3}}\). Then \(a^{7}\cdot b^{-3}\cdot c^{0}=a^{7}\cdot\frac{1}{b^{3}}\cdot1=\frac{a^{7}}{b^{3}}\).

Step4: Analyze the expression \(7b^{4}\cdot a^{-2}\cdot d^{6}\)

Using the negative exponent rule \(a^{-2}=\frac{1}{a^{2}}\), we have \(7b^{4}\cdot a^{-2}\cdot d^{6}=7b^{4}\cdot\frac{1}{a^{2}}\cdot d^{6}=\frac{7b^{4}d^{6}}{a^{2}}\).

Step5: Analyze the expression \(- 1\cdot k^{0}\cdot d^{-3}\cdot4d^{6}\)

Using the zero - exponent rule \(k^{0}=1\) (\(k
eq0\)) and the product rule of exponents \(a^{m}\cdot a^{n}=a^{m + n}\). First, \(k^{0}=1\), then \(-1\cdot1\cdot4\cdot d^{-3 + 6}=- 4d^{3}\)? Wait, there seems to be a mistake in the original writing. Let's recalculate: \(-1\cdot k^{0}\cdot d^{-3}\cdot4d^{6}=-1\times1\times4\times d^{-3 + 6}=-4d^{3}\). But the original has \(4d^{9}\), maybe there is a typo in the original problem.

Step6: Analyze the expression \((5\times10^{3})\cdot(0.2\times10^{-4})\)

Using the commutative and associative properties of multiplication: \((5\times0.2)\times(10^{3}\times10^{-4})\). Calculate \(5\times0.2 = 1\), and using the product rule of exponents \(a^{m}\cdot a^{n}=a^{m + n}\), \(10^{3}\times10^{-4}=10^{3-4}=10^{-1}=\frac{1}{10}\). So \((5\times10^{3})\cdot(0.2\times10^{-4})=1\times\frac{1}{10}=\frac{1}{10}\).

Step7: Analyze the expression \((4\times10^{0})\cdot(3\times10^{-2})\)

Using the zero - exponent rule \(10^{0}=1\) and the product rule of exponents. \(4\times3\times10^{0+( - 2)} = 12\times10^{-2}=\frac{12}{10^{2}}\) (since \(10^{-2}=\frac{1}{10^{2}}\)).

Step8: Analyze the expression \((2\times10^{2})\div(7\times10^{-7})\)

Using the quotient rule of exponents \(\frac{a^{m}}{a^{n}}=a^{m - n}\) and the division of coefficients. \(\frac{2}{7}\times\frac{10^{2}}{10^{-7}}=\frac{2}{7}\times10^{2-( - 7)}=\frac{2}{7}\times10^{9}\approx0.2857\times10^{9}=2.857\times10^{8}\), but the original has a mistake. If we consider the original's wrong step, maybe it's a miscalculation.

Answer:

The expressions involve the application of exponent rules (negative exponents, zero exponents, product and quotient rules of exponents). The correct simplifications (excluding the original mistakes) are as follows: \(2^{-8}=\frac{1}{256}\), \(16^{-2}=\frac{1}{256}\), \(a^{7}\cdot b^{-3}\cdot c^{0}=\frac{a^{7}}{b^{3}}\), \(7b^{4}\cdot a^{-2}\cdot d^{6}=\frac{7b^{4}d^{6}}{a^{2}}\), \((5\times10^{3})\cdot(0.2\times10^{-4})=\frac{1}{10}\), \((4\times10^{0})\cdot(3\times10^{-2})=\frac{12}{10^{2}}\)