I think Theodosius the best measure. The outliers will not effect the measure. If you use mean, you add all the data points and divide by the number of data points. The extremes will skew the data. If you use mode, that may work also. You find the data that occurs most often.
Answer:
If the absolute value expression is not equal to zero, the expression inside an absolute value can be either positive or negative. So, there can be at most two solutions. Looking at this graphically, an absolute value graph can intersect a horizontal line at most two times.
Answer:
Step-by-step explanation:
![\sqrt[3]{125y^9z^6}\\ \\ \sqrt[3]{5^3(y^3)^3(z^2)^3}\\ \\ 5y^3z^2](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B125y%5E9z%5E6%7D%5C%5C%20%5C%5C%20%5Csqrt%5B3%5D%7B5%5E3%28y%5E3%29%5E3%28z%5E2%29%5E3%7D%5C%5C%20%5C%5C%205y%5E3z%5E2)
In this equation, you have to treat the number in the bracket first on the basis of BODMAS
15 - [-3]- 4
Note that when two minuses come together the product is a plus sign.
15 +3 - 4
You have to add before you subract
18 - 4 =14
Therefore, 15- [-3] - 4 = 14.
Answer:
g(x) = (x + 6)^2 - 5
Step-by-step explanation:
To translate f(x) to the left by 6 units, replace x with x+6.
To translate f(x) down 5 units, replace f(x) with f(x)-5.
Together, these replacements give you ...
... g(x) = f(x+6) -5
... g(x) = (x +6)^2 -5
