11 (a) Find the value of \( \sqrt{81} \times 10^8 \)\n\n(b) Find the value of \( 64^{-\frac{1}{3}} \)\n\n(c) Write \( \frac{3^n}{9^{n-1}} \) as a power of 3 - Edexcel - GCSE Maths - Question 12 - 2020 - Paper 1
Question 12
11 (a) Find the value of \( \sqrt{81} \times 10^8 \)\n\n(b) Find the value of \( 64^{-\frac{1}{3}} \)\n\n(c) Write \( \frac{3^n}{9^{n-1}} \) as a power of 3
Worked Solution & Example Answer:11 (a) Find the value of \( \sqrt{81} \times 10^8 \)\n\n(b) Find the value of \( 64^{-\frac{1}{3}} \)\n\n(c) Write \( \frac{3^n}{9^{n-1}} \) as a power of 3 - Edexcel - GCSE Maths - Question 12 - 2020 - Paper 1
Step 1
Find the value of \( \sqrt{81} \times 10^8 \)
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Answer
To find ( \sqrt{81} ), we recognize that ( 81 = 9^2 ), thus ( \sqrt{81} = 9 ). Multiplying by ( 10^8 ):\n( 9 \times 10^8 = 9 \times 10^8 ). Therefore, the final answer is ( 9 \times 10^8 ).
Step 2
Find the value of \( 64^{-\frac{1}{3}} \)
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Answer
We start by finding the cube root of ( 64 ). Since ( 64 = 4^3 ), we have:\n( \sqrt[3]{64} = 4 ). Then, applying the negative exponent, we find:\n( 64^{-\frac{1}{3}} = \frac{1}{4} ). Thus, the answer is ( \frac{1}{4} ).
Step 3
Write \( \frac{3^n}{9^{n-1}} \) as a power of 3
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Answer
First, rewrite ( 9 ) as a power of 3: ( 9 = 3^2 ). Thus, we can rewrite the expression as:\n( \frac{3^n}{(3^2)^{n-1}} = \frac{3^n}{3^{2(n-1)}} = 3^{n - 2(n-1)} = 3^{n - 2n + 2} = 3^{2 - n} ). Therefore, the final expression as a power of 3 is ( 3^{2-n} ).
