Answer:
Step-by-step explanation:
From C-A, it goes from y=1 to y=5, so that 4 units
From A-B, it goes from x = -1 to x = 4, that is 5 units
Now, to find distance from B to C, we need to use the distance formula:
Where the variables are the respective points of B and C,
B (4,5) & C(-1,1)
So x_1 =4, y_1=5, x_2=-1, y_2=1
Plugging into the formula we get:
Summing it all (perimeter is sum of 3 sides):
Distance =
3rd answer choice is right.
a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
b. The inverse image of g(x) is
c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
<h3 /><h3>The domain of T</h3>Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0
⇒ x² ≥ 25
⇒ x ≥ ±5
⇒ -5 ≤ x ≤ 5.
<h3>Inverse image of f(x)</h3>
The inverse image of f(x) is f⁻¹(x) = ∛(x/3)
f : R → R defined by f(x) = 3x³
Let f(x) = y.
So, y = 3x³
Dividing through by 3, we have
y/3 = x³
Taking cube root of both sides, we have
x = ∛(y/3)
Replacing y with x we have
y = ∛(x/3)
Replacing y with f⁻¹(x), we have
So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)
<h3>Inverse image of g(x)</h3>
The inverse image of g(x) is
g : R+ → R defined by g(x) = ln(x).
Let g(x) = y
y = ln(x)
Taking exponents of both sides, we have
Replacing x with y, we have
Replacing y with g⁻¹(x), we have
So, the inverse image of g(x) is
The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
h : R → R defined by h(x) = x − 9
Let y = h(x)
y = x - 9
Adding 9 to both sides, we have
y + 9 = x
Replacing x with y, we have
x + 9 = y
Replacing y with h⁻¹(x), we have
So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9
Learn more about inverse image of a function here: