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3 years ago

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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3 years ago

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Answer:

Step-by-step explanation:

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3 years ago

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4 years ago

Which of the following tables represents a function? x 2 2 3 3 y 5 −5 6 −6 x 4 −4 4 −4 y 1 2 3 4 x 6 6 7 7 y 2 3 4 5 x 2 −2 4 −4

Crank

Answer: Fourth option.

Step-by-step explanation:

By definition, a relationship is a function if and only if each input value (or each x-value) has only one output value (or y-value).

Given the first table:

$x|2\ \ 2\ \ 3\ \ 3\\y|5-5\ 6-6$

You can notice that the input values 2 and 3 do not have only one output value. Therefore, this table does not represent a function.

The second table is:

$x|4-4\ 4\ -4\\y|1\.\ \ 2\ \.\ \ 3\ \ 4$

You can notice that the input values 4 and -4 do not have only one output value. Therefore, this table does not represent a function.

The third table is:

$x| 6\ 6\ 7\ 7\ \\y|2\ 3\ 4\ 5$

You can notice that the input values 6 and 7 do not have only one output value. Therefore, this table does not represent a function.

The fourth table is:

$x| 2\ -2\ 4\ -4\ \\y|8\ \.\ \ 8\ \ 12\ \ 12$

You can notice that each input value has only one output value. Therefore, this table represents a function.

4 0

4 years ago

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