Find the measure of the graph

kari74 [83]

Answer:

D) a 360° clockwise rotation about the center of the parallelogram will take the parallelogram PQRS onto itself.

Step-by-step explanation:

The parallelogram PQRS is in quadrant III (3) of the coordinate plane. If the image is reflected over the line x = -5 or y = -5, it would be the mirror image of itself and would move either left (x = -5) or down (y = -5) over the lines. A 180° rotation about the center of the parallelogram would move the object to quadrant I (1). However, a 360° clockwise rotation about the center of the parallelogram would cause the object to make a full circle and take the parallelogram PQRS onto itself.

5 0

3 years ago

Please help me on this one?

blondinia [14]

3x + 4y = 16 Subtract 4y from both sides. Put brackets around the 2 terms on the right.

3x + 4y - 4y = (16 - 4y)

3x = (16 - 4y) Divide by 3

$\text{ x =}\dfrac{(16 - 4y)}{3} = \dfrac{16}{3} - \dfrac{4y}{3}$

Answer That's B.

3 0

3 years ago

Read 2 more answers

Brianna writes 4×11 to show the number 44 as a product of its prime factors.Does her answer make sense? Explain your reasoning.

PilotLPTM [1.2K]

It does not make sense, because 4 is not a prime number. If it would have been 2*2*11 then it would have been correct.

3 0

3 years ago

Read 2 more answers

URGENT

Katyanochek1 [597]

Answer:

The answer to your question is Area = 62.5 u²

Step-by-step explanation:

Area = $c \frac{a + b}{2}$

a and b are bases

c = heigth

dAD = $\sqrt{(x2 - x1)^{2} - (y2 - y1)^{2} }$

dAD = $\sqrt{(-1 + 13)^{2} - (5 + 11)^{2} }$

dAD = $\sqrt{(12)^{2} - (16)^{2} }$

dAD = $\sqrt{144 + 256}$

dAD = $\sqrt{400}$

dAD = 20 u

dBC = $\sqrt{(3 - 0)^{2} - (2 + 2)^{2} }$

dBC = $\sqrt{9 + 16}$

dBC = $\sqrt{25}$

dBC = 5 u

dAB = $\sqrt{(3 + 1)^{2} - (2 - 5)^{2} }$

dAB = $\sqrt{(4)^{2} - (-3)^{2} }$

dAB = $\sqrt{16 + 9}$

dAB = $\sqrt{25}$

dAB = 5u

Area = $5 \frac{20 + 5}{2}$

Area = $5 \frac{25}{2}$

Area = 5(12.5)

Area = 62.5 u²

6 0

4 years ago

Describe the translation. y = (x-5)squared +5 - y = (x-))squared +0

vovangra [49]

In functions; translations are used to move lines, points or shapes across the graph (i.e. up, down, right or left).

The translation from $y = (x - 5)^2 +5$ to $y = (x - 0)^2 +0$ is to shift $y = (x - 5)^2 +5$ to the left by 5 units and then downward by 5 units

The given parameters are:

$y = (x - 5)^2 +5$

$y = (x - 0)^2 +0$

Assume the first point is: $y = (x - 5)^2 +5$

While the second is: $y = (x - 0)^2 +0$

The first point is first shifted left by 5 units

The rule to this translation is: $(x,y) \to (x + 5,y)$

So, we have:

$y' = (x - 5 + 5)^2 + 5$

$y' = (x - 0)^2 + 5$

Next, the resulting function is shifted down by 5 units.

The rule to this translation is: $(x,y) \to (x,y- 5)$

So, we have:

$y" = y' - 5$

This gives:

$y" = (x - 0)^2 + 5 - 5$

$y" = (x - 0)^2 + 0$

Hence, the translation from $y = (x - 5)^2 +5$ to $y = (x - 0)^2 +0$ is to shift $y = (x - 5)^2 +5$ to the left by 5 units and then downward by 5 units