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m_a_m_a [10]
3 years ago
7

Evaluate 1/ 2^-2x^-3y^5 for x=2 and y=-4

Mathematics
1 answer:
Aneli [31]3 years ago
6 0

Answer:

-512

Step-by-step explanation:

Evaluate y^5/(x^3/2^2) where x = 2 and y = -4:

(y^5)/(2^(-2) x^3) = (-4)^5/2^(-2)×1/2^3

2^3 = 2×2^2:

(-4)^5/2^(-2) 1/(2×2^2)

2^2 = 4:

(-4)^5/((2×4)/2^2)

2×4 = 8:

(-4)^5/(8/2^2)

2^(-2) = 1/4:

((-4)^5/8)/(1/4)

(-4)^5 = (-1)^5×4^5 = -4^5:

((-4^5)/8)/(1/4)

4^5 = 4×4^4 = 4 (4^2)^2:

((-4 (4^2)^2)/8)/(1/4)

4^2 = 16:

((-4×16^2)/8)/(1/4)

| 1 | 6

× | 1 | 6

| 9 | 6

1 | 6 | 0

2 | 5 | 6:

((-4×256)/8)/(1/4)

4×256 = 1024:

((-1024)/8)/(1/4)

Multiply the numerator by the reciprocal of the denominator, ((-1024)/8)/(1/4) = (-1024)/8×4:

(-1024×4)/8

(-1024)/8 = (8 (-128))/8 = -128:

-128×4

-128×4 = -512:

Answer:  -512

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