The coordinates of the vertex that A maps to after Daniel's reflections are (3, 4) and the coordinates of the vertex that A maps to after Zachary's reflections are (3, 2)
<h3>How to determine the coordinates of the vertex that A maps to after the two reflections?</h3>
From the given figure, the coordinate of the vertex A is represented as:
A = (-5, 2)
<u>The coordinates of the vertex that A maps to after Daniel's reflections</u>
The rule of reflection across the line x = -1 is
(x, y) ⇒ (-x - 2, y)
So, we have:
A' = (5 - 2, 2)
Evaluate the difference
A' = (3, 2)
The rule of reflection across the line y = 2 is
(x, y) ⇒ (x, -y + 4)
So, we have:
A'' = (3, -2 + 4)
Evaluate the difference
A'' = (3, 4)
Hence, the coordinates of the vertex that A maps to after Daniel's reflections are (3, 4)
<u>The coordinates of the vertex that A maps to after Zachary's reflections</u>
The rule of reflection across the line y = 2 is
(x, y) ⇒ (x, -y + 4)
So, we have:
A' = (-5, -2 + 4)
Evaluate the difference
A' = (-5, 2)
The rule of reflection across the line x = -1 is
(x, y) ⇒ (-x - 2, y)
So, we have:
A'' = (5 - 2, 2)
Evaluate the difference
A'' = (3, 2)
Hence, the coordinates of the vertex that A maps to after Zachary's reflections are (3, 2)
Read more about reflection at:
brainly.com/question/4289712
#SPJ1
Answer:
c
Step-by-step explanation:
because it's A.B
C.D.
Given,
Rate of free fall = 216 km/h
Solution,
km/h to m/s is converted as follows :
So, the rate of fall is 60 m/s.
If t = 20 s
Assume, initial velocity = 0
It will move under the action of gravity.
Using equation of motion,
Hence, this is all for the solution.
Answer:
x = 1
Step-by-step explanation:
4x-5=y
4x-5= -1
4x = -1+5
4x=4
x=1
2x-y=3
2x-(-1)=3
2x+1=3
2x= 3-1
2x= 2
x=1
since the two answers for x are the same therefore:
x = 1
PART A
s = <span>the number of packets of strawberry wafers ;
c = </span><span>the number of packets of chocolate wafers ;
3 </span>× s + 1 × <span>c = 30 ;
s + c = 22 ;
PART B
</span>The method of solving "by substitution"<span> works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable.
</span>
c = 30 - 3s;
s + ( 30 - 3s ) = 22;
30 - 2s = 22;
30 - 22 = 2s;
8 = 2s;
s = 4 ;
c = 30 - 12 ;
c = 18 ;
