Answer:
The domain of the inverse of a relation is the same as the range of the original relation. In other words, the y-values of the relation are the x-values of the inverse.
Thus, domain of f(x): x∈R = range of f¯¹(x)
and range of f(x): x∈R =domain of f¯¹(x)
Answer:
Step-by-step explanation:
In order to find the side mentioned (IJ), we need to use SOH CAH TOA.
SOH CAH TOA is an acronym to help us remember what sin, cos, and tan mean. It stands for:
Sin = Opposite / Hypotenuse
Cosine = Adjacent / Hypotenuse
Tan = Opposite / Adjacent
Since we know the measure of angle K (42) and we know one of the sides, we can use this to find the missing length.
Since the side given to us is the hypotenuse, and we're looking for the side opposite of the angle (IJ), the only possible one to use would be SIN as it includes Opposite and Hypotenuse.
Our equation is now this:
Let's now solve for x.
Therefore, the length of IJ will be around 2.01.
Hope this helped!
Step-by-step explanation:
In this case we have:
Δx = 3/n
b − a = 3
a = 1
b = 4
So the integral is:
∫₁⁴ √x dx
To evaluate the integral, we write the radical as an exponent.
∫₁⁴ x^½ dx
= ⅔ x^³/₂ + C |₁⁴
= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)
= ⅔ (8) + C − ⅔ − C
= 14/3
If ∫₁⁴ f(x) dx = e⁴ − e, then:
∫₁⁴ (2f(x) − 1) dx
= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx
= 2 (e⁴ − e) − (x + C) |₁⁴
= 2e⁴ − 2e − 3
∫ sec²(x/k) dx
k ∫ 1/k sec²(x/k) dx
k tan(x/k) + C
Evaluating between x=0 and x=π/2:
k tan(π/(2k)) + C − (k tan(0) + C)
k tan(π/(2k))
Setting this equal to k:
k tan(π/(2k)) = k
tan(π/(2k)) = 1
π/(2k) = π/4
1/(2k) = 1/4
2k = 4
k = 2
