To find inverse, you switch x and y, and solve for y
Original: y=2x-1/3
Now:
x=2y-1/3
x+1/3=2y
(x+1/3)/2=y
y=
What’s c? Not valid question
Using it's concept, it is found that the domain for the expressions is, respectively, given by:
![x \neq 0, x \neq 0, (-\infty, \infty), (-\infty, \infty), x \neq 2](https://tex.z-dn.net/?f=x%20%5Cneq%200%2C%20x%20%5Cneq%200%2C%20%28-%5Cinfty%2C%20%5Cinfty%29%2C%20%28-%5Cinfty%2C%20%5Cinfty%29%2C%20x%20%5Cneq%202)
<h3>What is the domain of a function?</h3>
It is the <u>set that contains all possible input values</u>.
In a fraction, the denominator cannot be zero, hence:
- The domain of the first two expressions is of
.
- The domain of the last expression is of
.
The third expression can be simplified, as:
(x + 5)/(x + 5) = 1.
The same is true for the fourth, as:
x²/x = 1.
Neither has any restriction, hence their domain is all real numbers, represented by
.
More can be learned about the domain of a function at brainly.com/question/25897115
Answer:
Here's what I find.
Step-by-step explanation:
You have 800 deer at the end of Year 1, and you expect the population to decrease each year thereafter.
a) i) The recursive formula
Let dₙ = the deer population n years after the initial measurement.
![d_{n} = d_{n - 1}r^{n}](https://tex.z-dn.net/?f=d_%7Bn%7D%20%3D%20d_%7Bn%20-%201%7Dr%5E%7Bn%7D)
For this situation,
![d_{n} = d_{n - 1}(0.5)^{n}](https://tex.z-dn.net/?f=d_%7Bn%7D%20%3D%20d_%7Bn%20-%201%7D%280.5%29%5E%7Bn%7D)
a) ii) Definitions
n = the number of years from first measurement
r = the common ratio, that is, the deer population at the end of one year divided by the population of the previous year.
a) iii) First term of sequence
The first term of the sequence is d₀, the population when first measured.
b) The function formula
The formula for the nth term of a geometric series is
![d_{n} =d_{0}r^{n - 1}](https://tex.z-dn.net/?f=d_%7Bn%7D%20%3Dd_%7B0%7Dr%5E%7Bn%20-%201%7D)
c) Value of d₀
Let n = 2; then d₂ = 800
![\begin{array}{rcl}800 & = & d_{0}(0.5)^{2 - 1}\\800 & = & d_{0}(0.5)\\\\d_{0} & = & \dfrac{800}{0.5}\\\\& =&\mathbf{1600}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7D800%20%26%20%3D%20%26%20d_%7B0%7D%280.5%29%5E%7B2%20-%201%7D%5C%5C800%20%26%20%3D%20%26%20d_%7B0%7D%280.5%29%5C%5C%5C%5Cd_%7B0%7D%20%26%20%3D%20%26%20%5Cdfrac%7B800%7D%7B0.5%7D%5C%5C%5C%5C%26%20%3D%26%5Cmathbf%7B1600%7D%5C%5C%5Cend%7Barray%7D)
Answer:
a=1 or a=5/6
Step-by-step explanation:
I'm going to attempt to factor 6a^2-a-5
a=6
b=-1
c=-5
Find two numbers that multiply to be a*c and add to be b.
a*c=-30 =-6(5)
b=-1 =-6+5
So replace -a with -6a+5a in the expression we started with
6a^2-6a+5a-5
now we factor by grouping
6a(a-1)+5(a-1)
(a-1)(6a-5)
Now let's solve the equation:
(a-1)(6a-5)=0
So a=1 or a=5/6