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3 years ago

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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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:

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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 $3(w-2)(6 w-30) =18(w-2)(w-5)$

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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)}$

Cancel the common factor (w – 2).

$$=-\frac{w+7}{9}$

$$\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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