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14 (a) Work out the value of \( \left( \frac{16}{81} \right)^{\frac{3}{2}} \) (b) Work out the value of \( a + b + c \) 3$^{-1}$ = 1/9 3$^{3}$ = 9$ adical{3}$ 3$^{\frac{3}{2}}$ = 1/$ adical{3}$ - Edexcel - GCSE Maths - Question 15 - 2018 - Paper 1
Question 15
14 (a) Work out the value of \( \left( \frac{16}{81} \right)^{\frac{3}{2}} \) (b) Work out the value of \( a + b + c \) 3$^{-1}$ = 1/9 3$^{3}$ = 9$ adica... show full transcript
Worked Solution & Example Answer:14 (a) Work out the value of \( \left( \frac{16}{81} \right)^{\frac{3}{2}} \) (b) Work out the value of \( a + b + c \) 3$^{-1}$ = 1/9 3$^{3}$ = 9$ adical{3}$ 3$^{\frac{3}{2}}$ = 1/$ adical{3}$ - Edexcel - GCSE Maths - Question 15 - 2018 - Paper 1
Step 1
Work out the value of \( \left( \frac{16}{81} \right)^{\frac{3}{2}} \)
Answer
To solve ( \left( \frac{16}{81} \right)^{\frac{3}{2}} ), we start by rewriting it in terms of square roots:
Calculating the square root first:
Now raising this result to the power of 3:
Thus, ( \left( \frac{16}{81} \right)^{\frac{3}{2}} = \frac{64}{729} ).
Step 2
Work out the value of \( a + b + c \)
Answer
From the equations given, let's express each in terms of powers of 3:
- For ( 3^{-1} = \frac{1}{9} ), we can write ( \frac{1}{9} = 3^{-2} ).
- For ( 3^{3} = 9 \sqrt{3} ), this can be rewritten as ( 9 \sqrt{3} = 3^{2} \cdot 3^{\frac{1}{2}} = 3^{2.5} = 3^{\frac{5}{2}} ).
- Finally, for ( 3^{\frac{3}{2}} = \frac{1}{\sqrt{3}} ), we know this rewrites as ( 3^{\frac{3}{2}} = 3^{1.5} ).
So we rewrite the addition:
Without specific values for ( a, b, c ), we cannot compute a numerical value, but we confirm they are in terms of powers of 3.