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3 years ago

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What are the coordinates of the image of the point (2, –6) under a dilation with a center of (0, 0) and a scale factor of 1/2 ?

A. (0, 0) B. (1, –3) C. (2, –6) D. (4, –12)

2 answers:

Troyanec [42]3 years ago

7 0

Answer:

B. (1, -3)

Step-by-step explanation:

When you dilate from the origin (0, 0) by a scale factor of 1/2, multiple each coordinate point by 1/2.

(2, -6)

2 × 1/2 = 1

-6 × 1/2 = -3

(1, -3)

I hope this helps :))

givi [52]3 years ago

4 0

Answer:

The answer is (1,-3)

Step-by-step explanation:

I took the test :)

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9w-4w+6<1+5w supposed to combine like terms

Mariana [72]

Answer/Step-by-step explanation:

Given:

$9w - 4w + 6$ < $1 + 5w$

To solve, collect like terms.

Subtract 5w from both sides

$9w - 4w + 6 - 5w$ < $1 + 5w - 5w$

$9w - 4w - 5w + 6$ < $1$

$6$ < $1$ (incorrect)

The inequality given has no solution or an information is missing.

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3 years ago

AB and BC form a right angle at point B. If A = (-3,-1) and B = (4,4), what is the equation of BC

m_a_m_a [10]

Answer:

$y=-1.4x+9.6$

Step-by-step explanation:

If A = (-3,-1) and B = (4,4), the slope of AB is

$\text{Slope}_{AB}=\dfrac{-1-4}{-3-4}=\dfrac{-5}{-7}=\dfrac{5}{7}$

Two perpendicular lines have slopes that have product of -1:

$\text{Slopee}_{AB}\cdot \text{Slope}_{BC}=-1\\ \\\dfrac{5}{7}\cdot \text{Slope}_{BC}=-1\\ \\\text{Slope}_{BC}=-\dfrac{7}{5}=-1.4$

The equation of the line BC with slope -1.4 and passing through the point B(4,4) is

$y-4=-1.4(x-4)\\ \\y-4=-1.4x+5.6\\ \\y=-1.4x+9.6$

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4 years ago

Expand<br> (3a – 5b)(2a +36)

olga_2 [115]

Answer:

6a^2+108a-10ab-180b

Step-by-step explanation:

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3 years ago

Point B has coordinates (4,1). The x-coordinate of point A is -4. The distance between point A and point B is 10 units. What a

klasskru [66]

Given:

Point B has coordinates (4,1).

The x-coordinate of point A is -4.

The distance between point A and point B is 10 units.

To find:

The possible coordinates of point A.

Solution:

Let the y-coordinate of point A be y. Then the two points are A(-4,y) and B(4,1).

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

The distance between point A and point B is 10 units.

$\sqrt{(4-(-4))^2+(1-y)^2}=10$

Taking square on both sides, we get

$(8)^2+(1-y)^2=100$

$(1-y)^2=100-64$

$(1-y)^2=36$

Taking square root on both sides, we get

$(1-y)=\pm \sqrt{36}$

$-y=\pm 6-1$

$y=1\mp 6$

$y=1-6$ and $y=1+6$

$y=-5$ and $y=7$

Therefore, the possible coordinates of point A are either (-4,-5) or (-4,7).

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3 years ago

Point (-1,3) lies in which quadrant ?

777dan777 [17]

Quadrant II

Or the top left

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2 years ago

Read 2 more answers