Answer:
When we have a rational function like:
The domain will be the set of all real numbers, such that the denominator is different than zero.
So the first step is to find the values of x such that the denominator (x^2 + 3) is equal to zero.
Then we need to solve:
x^2 + 3 = 0
x^2 = -3
x = √(-3)
This is the square root of a negative number, then this is a complex number.
This means that there is no real number such that x^2 + 3 is equal to zero, then if x can only be a real number, we will never have the denominator equal to zero, so the domain will be the set of all real numbers.
D: x ∈ R.
b) we want to find two different numbers x such that:
r(x) = 1/4
Then we need to solve:
We can multiply both sides by (x^2 + 3)
Now we can multiply both sides by 4:
Now we only need to solve the quadratic equation:
x^2 + 3 - 4*x - 4 = 0
x^2 - 4*x - 1 = 0
We can use the Bhaskara's formula to solve this, remember that for an equation like:
a*x^2 + b*x + c = 0
the solutions are:
here we have:
a = 1
b = -4
c = -1
Then in this case the solutions are:
x = (4 + 4.47)/2 = 4.235
x = (4 - 4.47)/2 = -0.235
Answer:
Kindly check explanation
Step-by-step explanation:
The following mistakes could be explained observed from the plotted bar chart presented in the attachment :
1.) The numeric label of the y-axis which shows the number of text messages isn't well scaled, the numbering did not start from 0, hence this is a defect in the scaling of the bar chart.
2.) Looking closely, the plotted bars do not ALIGN well in terms of spacing as the interval between each bar isn't consistent.
3.) Both the x and y axis of the graph are t labeled thus failing to add a good and meaningful description to the plot especially for first time viewers.
Answer:
x + 3
Step-by-step explanation:
It cuts the y axis at 3 and continues to rise by one
Therefore ax + b gives us 1x + 3 = x + 3
Answer:
3r(to the secondpower) + 20r - 32
Step-by-step explanation:
This is a strange question, and f(x) may not even exist. Why do I say that? Well..
[1] We know that f(a+b) = f(a) + f(b). Therefore, f(0+0) = f(0) + f(0). In other words, f(0) = f(0) + f(0). Subtracting, we see, f(0) - f(0) = f(0) or 0 = f(0).
[2] So, what's the problem? We found the answer, f(0) = 0, right? Maybe, but the second rule says that f(x) is always positive. However, f(0) = 0 is not positive!
Since there is a contradiction, we must either conclude that the single value f(0) does not exist, or that the entire function f(x) does not exist.
To fix this, we could instead say that "f(x) is always nonnegative" and then we would be safe.
