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DENIUS [597]
2 years ago
12

Find the perimeter of ABC with vertices A (1,1), B (7,1), and C (1,9)

Mathematics
1 answer:
Effectus [21]2 years ago
4 0

Answer:

24 unit

Step-by-step explanation:

Given,

The vertices of the triangle ABC are,

A (1,1), B (7,1), and C (1,9),

By the distance formula,

AB=\sqrt{(7-1)^2+(1-1)^2}=\sqrt{6^2}=6\text{ unit}

BC=\sqrt{(1-7)^2+(9-1)^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{ unit}

CA=\sqrt{(1-1)^2+(1-9)^2}=\sqrt{8^2}=8\text{ unit}

Thus, the perimeter of the triangle ABC = AB + BC + CA = 6 + 10 + 8 = 24 unit

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Solve for b in the literal equation y = 11x + 11b
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<h2>b = y/11  -x </h2>

Step-by-step explanation:

y = 11x + 11b\\Move \: 11x\:to\:the \:left\:and\:change\:its\:sign\\\\y -11x =11b\\\\Divide\:both\:sides\:of\:the\:equation\:by\: 11\\\frac{y}{11} -\frac{11x}{11} = \frac{11b}{11} \\y/11 -x =b\\\\b = \frac{y}{11} - x

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What is 54-36÷9+14=n what is the value in n?
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Which describes how square S could be transformed to square S prime in two steps? Assume that the center of dilation is the orig
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Answer:

The correct option is;

A dilation by a scale factor of Two-fifths and then a translation of 3 units up

Step-by-step explanation:

Given that the coordinates of the vertices of square S are

(0, 0), (5, 0), (5, -5), (0, -5)

Given that the coordinates of the vertices of square S' are

(0, 1), (0, 3), (2, 3), (2, 1)

We have;

Length of side, s, for square S is s = √((y₂ - y₁)² + (x₂ - x₁)²)

Where;

(x₁, y₁) and (x₂, y₂) are the coordinates of two consecutive vertices

When (x₁, y₁) = (0, 0) and (x₂, y₂) = (5, 0), we have;

s = √((y₂ - y₁)² + (x₂ - x₁)²) = s₁ = √((0 - 0)² + (5 - 0)²) = √(5)² = 5

For square S', where (x₁, y₁) = (0, 1) and (x₂, y₂) = (0, 3)

Length of side, s₂, for square S' is s₂ = √((3 - 1)² + (0 - 0)²) = √(2)² = 2

Therefore;

The transformation of square S to S' involves a dilation of s₂/s₁ = 2/5

The after the dilation (about the origin),  the coordinates of S becomes;

(0, 0) transformed to (remains at) (0, 0) ....center of dilation

(5, 0) transformed to (5×2/5, 0) = (2, 0)

(5, -5) transformed to (2, -2)

(0, -5) transformed to (0, -2)

Comparison of (0, 0), (2, 0), (2, -2), (0, -2) and (0, 1), (0, 3), (2, 3), (2, 1) shows that the orientation is the same;

For (0, 0), (2, 0), (2, -2), (0, -2) we have;

(0, 0), (2, 0) the same y-values, (∴parallel to the x-axis)

(2, -2), (0, -2) the same y-values, (∴parallel to the x-axis)

For (0, 1), (0, 3), (2, 3), (2, 1) we have;

(0, 3), (2, 3) the same y-values, (∴parallel to the x-axis)

(0, 1), (2, 1) the same y-values, (∴parallel to the x-axis)

Therefore, the lowermost point closest to the y-axis in (0, 0), (2, 0), (2, -2), (0, -2) which is (0, -2) is translated to the lowermost point closest to the y-axis in (0, 1), (0, 3), (2, 3), (2, 1) which is (0, 1)

That is (0, -2) is translated to (0, 1) which shows that the translation is T((0 - 0), (1 - (-2)) = T(0, 3) or 3 units up

The correct option is therefore a dilation by a scale factor of Two-fifths and then a translation of 3 units up.

7 1
2 years ago
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