9514 1404 393
Answer:
- R'(-2, 2)
- F'(2, 2)
- G'(-2, -2)
Step-by-step explanation:
It can be useful to keep a list of the 90° rotation transformations.
(x, y) ⇒ (-y, x) . . . . . . 90° CCW, 270° CW
(x, y) ⇒ (-x, -y) . . . . . . 180°
(x, y) ⇒ (y, -x) . . . . . . . 270° CCW, 90° CW
__
1) (x, y) ⇒ (-x, -y) . . . . 180°
R(2, -2) ⇒ R'(-2, 2)
__
2) (x, y) ⇒ (-y, x) . . . . 90°
F(2, -2) ⇒ F'( 2, 2)
__
3) (x, y) ⇒ (y, -x) . . . . 270°
G(2, -2) ⇒ G'(-2, -2)
Answer:
C ≤ 2
Step-by-step explanation:
Given that :
Clayton is at most 2 meters above sea level ;
Let Clayton = C
Clayton's distance can be represented by the inequality :
2 meters above sea level is positive, +2
At most 2 meters above sea level means, Clayton could be anywhere between + 2 metwrs but not more than 2 meters
Hence, Clayton's distance can be represented by the inequality :
C ≤ 2
Answer:
y - 3 = ⁵/₂(x - 2) {the point-slope form of the equation}
y = ⁵/₂x - 2 {the slope-intercept form of the equation}
5x - 2y = 4 {the standard form of the equation}
Step-by-step explanation:
The point-slope form of equation is: y - y₀ = m(x - x₀), where (x₀, y₀) is any point the line passes through and m is the slope.
m = ⁵/₂
(2, 3) ⇒ x₀ = 2, y₀ = 3
So, the point-slope form of the equation:
y - 3 = ⁵/₂(x - 2)
Therefore:
y - 3 = ⁵/₂x - 5 {add 3 to both sides}
y = ⁵/₂x - 2 ← the slope-intercept form of the equation
-⁵/₂x + y = - 2 {multiply both sides by (-2)}
5x - 2y = 4 ← the standard form of the equation
It’s 0.0000805 (hope this helps!)
Step-by-step explanation:
5. Let the equation of the line be y = mx + c
m = (6-15)/(2-(-1)) = -3
sub (2, 6) and m = -3:
6 = -3(2) + c
c = 12
Equation of the line: y = -3x + 12
6. Let the equation of the line be y = mx + c
y-intercept of 7 means c = 7
y = mx + 7
x-intercept: value of x when y = 0
therefore, sub (4, 0):
0 = m(4) + 7
m = -7/4
Equation of the line: y = -7/4x + 7
this is a topic on coordinate geometry. If you wish to explore more into this topic you can give me a follow on Instagram (learntionary) I'll be posting the notes for this topic soon and will also be posting other topics' notes and some tips :)
