√4 + <span>3√9 = 2+3*3 = 11.
This is a rational number, since the answer can be express as a fraction (e.g., 11/1).
This does not hold for square roots in general. For example </span>√3 is irrational.
Answer: The company currently has 17 employees
Step-by-step explanation:
Let X represent the number of current employees in the company
From the first information, it can be expressed mathematically as
X + 3 ≥ 20
X ≥ 20 - 3
X ≥ 17
From the second information, it can be expressed mathematically as
X - 3 < 15
X < 15 + 3
X < 18
From the above solutions, it can be deduced that
17 ≤ X < 18
The only number that fulfils this criteria is 17.
Therefore, X = 17
Answer:
y = (-5/3)x - 4
Explanation:
The equation of a line with slope m that goes through the point (x1, y1) is:

Additionally, two lines are parallel if they have the same slope.
The slope of the line y = (-5/3)x - 2 is (-5/3) because it is the number beside x.
So, the slope of our line is also -5/3.
Therefore, replacing m by -5/3 and (x1, y1) by (-3, 1), we get that the equation of the line is:

Finally, solving for y, we get:

So, the answer is:
y = (-5/3)x - 4
Find the area of the small circle then the get the radius of the big circle add both radius' (12) . you would use the normal formula to find the area of the big circle as well . you should get your answer that way !
The solution set 0 and 8 are the true values of the absolute value equation
The absolute value equation that has the a solution set of 0 and 6 is |x - 3| = 3
<h3>How to determine the absolute value equation?</h3>
From the figure, the solution sets of the absolute value equation are given as:
x = {0, 6}
Calculate the mean of the solution set
Mean = 0.5 * (0 + 6)
Mean = 3
Calculate the difference of the solution set divided by 2
Difference = (6 - 0)/2
Difference = 3
The absolute value equation is the represented as:
|x - Mean| = Difference
Substitute known values
|x - 3| = 3
Hence, the absolute value equation that has the a solution set of 0 and 6 is |x - 3| = 3
Read more about absolute value equation at:
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