A comparison between a function and its inverse would show that the domain and range of the original function swap. The domain of the function becomes the range of the inverse, the range of the function becomes the domain of its inverse.
Looking at ordered pairs of the function and its inverse would look like this:
(2,4) on the original function becomes (4,2) on the inverse.
While the graph of a function and its inverse are noticeably different an important thing to note is that it is merely a reflection across the line y=x.
So even though they appear different you are looking at the same relationship just as y vs. x instead of x vs. y
Answer:
Investment A = 3650
Investment B = 1350
Step-by-step explanation:
Given that:
Total interest earned = 332.50
Total Principal invested = 5000
Investment A:
Rate = 8%
Time = 1 year
Principal = a
Interest earned = 0.08a
Investment B:
Rate = 3%
Time = 1 year
Principal = b
Interest earned = 0.03b
a + b = 5000 ____(1)
0.08a + 0.03b = 332.50 - - - (2)
From (1)
a = 5000 - b
Into (2)
0.08(5000 - b) + 0.03b = 332.50
400 - 0.08b + 0.03b = 332.50
400 - 0.05b = 332.50
-0.05b = - 67.5
b = $1350
a = 5000 - b
a = 5000 - 1350
a = $3650
Goran completes 6. karen completes 10. it takes 60 minutes
A
B
Just search it up on a math website or sum
Answer:
[(x + 6), (y + 1)]
Step-by-step explanation:
Vertices of the quadrilateral ABCD are,
A → (-5, 2)
B → (-3, 4)
C → (-2, 4)
D → (-1, 2)
By reflecting the given quadrilateral ABCD across x-axis to form the image quadrilateral A'B'C'D',
Rule for the reflection of a point across x-axis is,
(x, y) → (x , -y)
Coordinates of the image point A' will be,
A(-5, 2) → A'(-5, -2)
From the picture attached, point E is obtained by translation of point A'.
Rule for the translation of a point by h units right and k units up,
A'(x+h, y+k) → E(x', y')
By this rule,
A'(-5 + h, -2 + k) → E(1, -1)
By comparing coordinates of A' and E,
-5 + h = 1
h = 6
-2 + k = -1
k = 1
That means
Rule for the translation will be,
[(x + 6), (y + 1)]
