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Harman [31]
3 years ago
6

Figure ABCD is a rhombus with point A (3, −1). What rule would rotate the figure 270° counterclockwise?

Mathematics
2 answers:
Radda [10]3 years ago
4 0

Answer:

The correct option is 4.

Step-by-step explanation:

Given information: ABCD is a rhombus with point A (3, −1).

If a figure rotated 90° counterclockwise, then

(x,y)\rightarrow (-y,x)

If a figure rotated 180° counterclockwise, then

(x,y)\rightarrow (-x,-y)

If a figure rotated 270° counterclockwise, then

(x,y)\rightarrow (y,-x)

Therefore the rule (x,y)\rightarrow (y,-x) would rotate the figure 270° counterclockwise.

The correct option is 4.

hoa [83]3 years ago
3 0
(x,y)→(y,−x)
.......
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$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}

Solution:

Given expression:

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}

To solve the given expression:

First simplify: \frac{50-2 w^{2}}{3 w^{2}+9 w-30}

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30}=-\frac{2(w+5)(w-5)}{3(w-2)(w+5)}

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Now substitute this in the given expression.

$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{2(w-5)}{3(w-2)} \cdot \frac{w^{2}+5 w-14}{6 w-30}

Multiply the fractions \frac{a}{b} \cdot \frac{c}{d}=\frac{a \cdot c}{b \cdot d}

                               $=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{3(w-2)(6 w-30)}

Factor the denominator 3(w-2)(6 w-30) =18(w-2)(w-5)

                               $=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{18(w-2)(w-5)}                        

Cancel the common factor 2(w – 5).

                               $=-\frac{w^{2}+5 w-14}{9(w-2)}

Factor the numerator w^{2}+5 w-14=(w-2)(w+7)

                               $=-\frac{(w-2)(w+7)}{9(w-2)}

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$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}

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