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3 years ago

20. What is the transformation?:

Mathematics

2 answers:

schepotkina [342]3 years ago

8 0

Answer:

Reflection

Step-by-step explanation:

A rotation would be to rotate a figure around the origin

A reflection would be to reflect the figure around one of the axes.

A translation would be to move a figure a certain amount of units.

Since A and A' are 2 units away from the y-axis, and C, B, B', and C' are all one unit away from the y-axis, this could not be a translation because the position of A'B'C' is not the same position of ABC.

It could not be a rotation around the origin or any point because it would result the figure in either Quadrant 2 or 3, and in the one occasion it would be in Quadrant 1, the figure cannot be in that position.

This reveals only one option, that which is a reflection. A reflection about the x-axis would not make sense, since it would result in Quadrant 3, so a reflection around the y-axis would make the most sense.

The data we have above also accounts for a reflection, since all points are a certain distant away from the y-axis.

zheka24 [161]3 years ago

7 0

Reflection ,, it just made a copy of the shape on the other side , it’s identical which means it’s a reflection

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The correct answer is: " 1 m " .
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3 years ago

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Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

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Make h the subject:

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$x^2 = a^2 - h^2$ becomes

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$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

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$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

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$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

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3 years ago

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Svet_ta [14]

Answer:

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Step-by-step explanation:

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