Answer:
- A. segment A double prime B double prime = segment AB over 2
Step-by-step explanation:
<u>Triangle ABC with coordinates of:</u>
- A = (-3, 3), B = (1, -3), C = (-3, -3)
<u>Translation (x + 2, y + 0), coordinates will be:</u>
- A' = (-1, 3), B = ( 3, -3), C = (-1, -3)
<u>Dilation by a scale factor of 1/2 from the origin, coordinates will be:</u>
- A'' = (-0.5, 1.5), B'' = (1.5, -1.5), C= (-0.5, -1.5)
<u>Let's find the length of AB and A''B'' using distance formula</u>
- d = √(x2-x1)² + (y2 - y1)²
- AB = √(1-(-3))² + (-3 -3)² = √4²+6² = √16+36 = √52 = 2√13
- A''B'' = √(1.5 - (-0.5)) + (-1.5 - 1.5)² = √2²+3² = √13
<u>We see that </u>
<u>Now the answer options:</u>
A. segment A double prime B double prime = segment AB over 2
B. segment AB = segment A double prime B double prime over 2
- Incorrect. Should be AB = A''B''*2
C. segment AB over segment A double prime B double prime = one half
- Incorrect. Should be AB/A''B'' = 2
D. segment A double prime B double prime over segment AB = 2
- Incorrect. Should be A''B''/AB = 1/2
<span>I have graphed the given coordinates of both pentagons. The reflection was across the y-axis where coordinates of Pentagon PQRST (-x,y) resulted to Pentagon P'Q'R'S'T' (x,y). The line of reflection between the pentagons was x = 0. Line of reflection is the midway between the pre-image and its reflection.</span>
(1) R+G+B=50
(2) R=B+6
(3) G=B-4 so substitute G in (1)
R+B-4+B=50 substitute (2) in here
B+6+B-4+B=50 group like terms and solve
3B+2=50
3B=50-2
B=48/3=16 substitute in (2) and (3)
R=16+6=22
G=16-4=12
Check
22+12+16=50✅
I think the answer is 0
8*3=24
24 *0=0
Answer:
Step-by-step explanation:
The general equation for a circle is
Where a and b are the x and y values of the centre. So comparing to your equation you can see that a is 2/3 and b is 0
