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ch4aika [34]
3 years ago
10

Evaluate, when x=2 y=-5 and z=3

Mathematics
1 answer:
gregori [183]3 years ago
8 0

Answer:

-139

Step-by-step explanation:

Evaluate 1/4 (4 x^3 - 2 y - 2 z^3) y^2 - 16 x^2 where x = 2, y = -5 and z = 3:

(4 x^3 - 2 y - 2 z^3)/4 y^2 - 16 x^2 = (4×2^3 - -5×2 - 2×3^3)/4×(-5)^2 - 16×2^2

(4×2^3 - 2 (-5) - 2×3^3)/4×(-5)^2 = ((4×2^3 - 2 (-5) - 2×3^3) (-5)^2)/4:

((4×2^3 - 2 (-5) - 2×3^3) (-5)^2)/4 - 16×2^2

(-5)^2 = 25:

((4×2^3 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2^3 = 2×2^2:

((4×2×2^2 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2^2 = 4:

((4×2×4 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

2×4 = 8:

((4×8 - 2 (-5) - 2×3^3) 25)/4 - 16×2^2

3^3 = 3×3^2:

((4×8 - 2 (-5) - 23×3^2) 25)/4 - 16×2^2

3^2 = 9:

((4×8 - 2 (-5) - 2×3×9) 25)/4 - 16×2^2

3×9 = 27:

((4×8 - 2 (-5) - 227) 25)/4 - 16×2^2

4×8 = 32:

((32 - 2 (-5) - 2×27) 25)/4 - 16×2^2

-2 (-5) = 10:

((32 + 10 - 2×27) 25)/4 - 16×2^2

-2×27 = -54:

((32 + 10 + -54) 25)/4 - 16×2^2

| 3 | 2

+ | 1 | 0

| 4 | 2:

(42 - 54 25)/4 - 16×2^2

42 - 54 = -(54 - 42):

(-(54 - 42) 25)/4 - 16×2^2

| 5 | 4

- | 4 | 2

| 1 | 2:

(-12×25)/4 - 16×2^2

(-12)/4 = (4 (-3))/4 = -3:

-3×25 - 16×2^2

2^2 = 4:

-3×25 - 164

-3×25 = -75:

-75 - 16×4

-16×4 = -64:

-64 - 75

-75 - 64 = -(75 + 64):

-(75 + 64)

| 7 | 5

+ | 6 | 4

1 | 3 | 9:

Answer:  -139

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\sf\huge\underline\red{SOLUTION:}

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If MP is a perpendicular bisector to LN, then NP and LP are equivalent.

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