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Gre4nikov [31]
3 years ago
10

Which red triangle shows a 90° counterclockwise rotation of the blue triangle? Check all that apply. On a coordinate plane, a bl

ue triangle has points (negative 1, 4), (negative 5, 4), (negative 1, 1). A red triangle has points (negative 4, negative 1), (negative 1, negative 1), (negative 4, negative 5). On a coordinate plane, a blue triangle has points (1, 1), (4, 5), (4, 1). A red triangle has points (negative 1, 1), (negative 1, 4), (negative 5, 4). On a coordinate plane, a blue triangle has points (0, 1), (4, 4), (4, 1). A red triangle has points (negative 4, 1), (negative 4, 4), (0, 1). On a coordinate plane, a blue triangle has points (negative 5, 4), (negative 1, 4), (negative 1), 1). A red triangle has points (1, 1), (4, 1), (4, 5). On a coordinate plane, a blue triangle has points (negative 2, 5), (4, 5), (4, 1). A red triangle has points (negative 1, 4), (negative 5, 4), (negative 5, negative 2).
Mathematics
2 answers:
Tcecarenko [31]3 years ago
8 0

Question:

1) Blue triangle points; (-1, 4), (-5, 4), and (-1, 1)

Red triangle points; (-4, -1), (-1, -1). and (-4, -5)

2) Blue triangle points (1, 1), (4, 5), (4, 1)

Red triangle points; (-1, 1), (-1, 4). and (-5, 4)

3) Blue triangle points (0, 1), (4, 4), (4, 1)

Red triangle points; (-4, 1), (-4, 4). and (0, 1)

4) Blue triangle points (-5, 4), (-1, 4), (-1, 1)

Red triangle points; (1, 1), (4, 1). and (4, 5)

5) Blue triangle points (-2, 5), (4, 5), (4, 1)

Red triangle points; (-1, 4), (-5, 4). and (-5, -2)

Answer:

The correctly rotated red triangles are those of (1), (2), and (5)

Step-by-step explanation:

In a 90° counterclockwise rotation, every x and y points of the original triangle are switched while the y is turned negative, as shown in the following equation;

(x, y) to (-y, x)

Therefore, the triangles that undergo a 90° counterclockwise rotation are as follows;

1) Blue triangle points; (-1, 4), (-5, 4), and (-1, 1)

Red triangle points; (-4, -1), (-1, -1). and (-4, -5)

(-1, 4) → (-4, -1)

(-5, 4) → (-4, -5)

(-1, 1) → (-1, -1)

Correctly rotated 90° counterclockwise

2) Blue triangle points (1, 1), (4, 5), (4, 1)

Red triangle points; (-1, 1), (-1, 4). and (-5, 4)

(1, 1) → (-1, 1)

(4, 5) → (-5, 4)

(4, 1) → (-1, 4)

Correctly rotated 90° counterclockwise

3) Blue triangle points (0, 1), (4, 4), (4, 1)

Red triangle points; (-4, 1), (-4, 4). and (0, 1)

(0, 1) → (0, 1) ≠ (-1, 0)

(4, 4) → (-4, 4)

(4, 1) → (-4, 1)

Not correctly rotated 90° counterclockwise

4) Blue triangle points (-5, 4), (-1, 4), (-1, 1)

Red triangle points; (1, 1), (4, 1). and (4, 5)

(-5, 4) → (4, 5) ≠ (-4, -5)

(-1, 4) → (4, 1)

(-1, 1) → (1, 1)

Not correctly rotated 90° counterclockwise

5) Blue triangle points (-2, 5), (4, 5), (4, 1)

Red triangle points; (-1, 4), (-5, 4). and (-5, -2)

(-2, 5) → (-5, -2)

(4, 5) → (-5, 4)

(4, 1) → (-1, 4)

Correctly rotated 90° counterclockwise.

aev [14]3 years ago
7 0

Answer: A, B, E

Step-by-step explanation: On Edgenuity!!!

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Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as e=\frac{c}{a}

a) Eccentricity close to 0

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\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)

b) Eccentricity =5

5=\frac{c}{a} \:c=5a

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

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a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)

Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1

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What is a solution for <br> 72+12x=24+24x<br> 48=12x<br> 4=x
Wewaii [24]
72+12x=24+24x
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Step-by-step explanation:

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