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belka [17]
2 years ago
6

How is copying segment similar to copying an angle ?

Mathematics
1 answer:
pochemuha2 years ago
3 0

Answer:

Copying a line segment or angle is an illustration of transformation. Both are similar, because they are both rigid transformation.

When an angle is copied to a new angle, the new angle and the old angle will have the same measure

Similarly;

When a line segment is copied to a new segment, the new line segment and the old line segment will have the same measure

This type of transformation is referred to as a rigid transformation.

Hence, copying a line segment is similar to copying an angle, because they are both rigid transformations

Step-by-step explanation:

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Find dy/dx by implicit differentiation.
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dy/dx by implicit differentiation is cos(πx)/sin(πy)

<h3>How to find dy/dx by implicit differentiation?</h3>

Since we have the equation

(sin(πx) + cos(πy)⁸ = 17, to find dy/dx, we differentiate implicitly.

So, [(sin(πx) + cos(πy)⁸ = 17]

d[(sin(πx) + cos(πy)⁸]/dx = d17/dx

d[(sin(πx) + cos(πy)⁸]/dx = 0

Let sin(πx) + cos(πy) = u

So, du⁸/dx = 0

du⁸/du × du/dx = 0

Since,

  • du⁸/du = 8u⁷ and
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= dsin(πx)/dx + dcos(πy)/dx

= dsin(πx)/dx + (dcos(πy)/dy × dy/dx)

= πcos(πx) - πsin(πy) × dy/dx

So, du⁸/dx = 0

du⁸/du × du/dx = 0

8u⁷ × [ πcos(πx) - πsin(πy) × dy/dx] = 0

8[(sin(πx) + cos(πy)]⁷ ×  (πcos(πx) - πsin(πy) × dy/dx) = 0

Since 8[(sin(πx) + cos(πy)]⁷ ≠ 0

(πcos(πx) - πsin(πy) × dy/dx) = 0

πcos(πx) = πsin(πy) × dy/dx

dy/dx = πcos(πx)/πsin(πy)

dy/dx = cos(πx)/sin(πy)

So, dy/dx by implicit differentiation is cos(πx)/sin(πy)

Learn more about implicit differentiation here:

brainly.com/question/25081524

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