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Ksenya-84 [330]
3 years ago
13

The coordinates of one vertex of this triangle are (–2, 2). If the triangle is translated 3 units to the left and 5 units down,

what are the coordinates of the image of this point?
Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
7 0

Answer: (-5,-3)

Step-by-step explanation:

first you must know left and right is the x-axis while you and down is 4-axis, down and left is subtraction, right and up is addition

so 3 units left would affect x axis therefor subtracting 3 from -2 which is -5

now 5 units down would be the y-axis and subtracting, 2-5= -3

finally your answer will be (-5,-3)

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Write an equation of a line in point-slope and slope intercept form that passes through the point (1, -8) and is perpendicular t
prisoha [69]

Answer:

11y-2x= -90

Step-by-step explanation:

for the point (1,-8)

x=1 and y= -8

for the equation,

2x -4y= -11

x= -1/m

m= -1/x

x = -1/ (-11/2 )

m= 2/11

so for the equation,

y-y1= m(x- x1)

y- -8=2/11(x-1)

11(y+8)= 2(x-1)

11y +88= 2x-2

11y-2x= -90

3 0
3 years ago
IF k/3- 9 = 12, what is the value of k?<br> Α. 1<br> B. 7 <br> C. 9<br> D.63
Mkey [24]

Answer:

D. 63

Step-by-step explanation:

k/3 - 9 = 12

k/3 = 21

k = 63

6 0
2 years ago
Read 2 more answers
Line j passes through points (10, 10) and (1, 3). Line k is parallel to line j. What is the slope of line k?
SOVA2 [1]

Answer:

= (10-3) ÷ (10-1) = 7 ÷9 = 7/9

Step-by-step explanation:

As they are parallel, their slopes are the same. So if we find slope of line j we find slope of line k when we have two points of one line like (x1 , y1) and (x2 , y2) the slope of that line is (y2-y1) ÷ (x2-x1)

6 0
3 years ago
Solve for Literal equation for y<br> 5 = 4x+8y
amid [387]
1/8=y is the answer
6 0
3 years ago
1. In circle o, chord AB and tangent CAD are drawn. It is known that mAB=116 . Find each of the
meriva

Given:

m(ar AB) = 116°

To find:

1. The measure of arc BEA.

2. The measure of ∠CAB

3. The measure of ∠BAD

Solution:

1. The measure of arc of full circle is 360°.

m(ar AB) + m(ar BEA) = 360°

116° + m(ar BEA) = 360°

Subtract 116° from both sides.

m(ar BEA) = 244°

2. Using the tangent-chord angle theorem:

$\Rightarrow m\angle CAB = \frac{1}{2} m(ar \ AB)

$\Rightarrow m\angle CAB = \frac{1}{2} \times 116^\circ

⇒ m∠CAB = 58°

3. Using the tangent-chord angle theorem:

$\Rightarrow m\angle BAD = \frac{1}{2} m(ar \ BEA)

$\Rightarrow m\angle BAD = \frac{1}{2} \times 244^\circ

⇒ m∠BAD = 122°

Using the fact that it is supplementary to ∠CAB.

⇒ m∠CAB + m∠BAD = 180°

⇒ 58° + m∠BAD = 180°

Subtract 58° from both sides.

⇒ m∠BAD = 122°

4 0
3 years ago
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