Answer:
In inequality notation:
Domain: -1 ≤ x ≤ 3
Range: -4 ≤ x ≤ 0
In set-builder notation:
Domain: {x | -1 ≤ x ≤ 3 }
Range: {y | -4 ≤ x ≤ 0 }
In interval notation:
Domain: [-1, 3]
Range: [-4, 0]
Step-by-step explanation:
The domain is all the x-values of a relation.
The range is all the y-values of a relation.
In this example, we have an equation of a circle.
To find the domain of a relation, think about all the x-values the relation can be. In this example, the x-values of the relation start at the -1 line and end at the 3 line. The same can be said for the range, for the y-values of the relation start at the -4 line and end at the 0 line.
But what should our notation be? There are three ways to notate domain and range.
Inequality notation is the first notation you learn when dealing with problems like these. You would use an inequality to describe the values of x and y.
In inequality notation:
Domain: -1 ≤ x ≤ 3
Range: -4 ≤ x ≤ 0
Set-builder notation is VERY similar to inequality notation except for the fact that it has brackets and the variable in question.
In set-builder notation:
Domain: {x | -1 ≤ x ≤ 3 }
Range: {y | -4 ≤ x ≤ 0 }
Interval notation is another way of identifying domain and range. It is the idea of using the number lines of the inequalities of the domain and range, just in algebriac form. Note that [ and ] represent ≤ and ≥, while ( and ) represent < and >.
In interval notation:
Domain: [-1, 3]
Range: [-4, 0]
Jacob is correct in thinking that because the mean represents the average of several quantities
C = 9/28 = 0.321
https://www.tiger-algebra.com/drill/3c-3/4=2c/3/
Answer:
Again! Option D is the answer!
Please mark my answer as the brainliest since I have you the answer so quickly!
This is a right triangle, so first find the distance between the two legs...
Points where the numbers are the same indicate a point that is on the same axis as the other
(-3, 3) and (-3, 2) have a distance of 1
(-3,2) and (1,2) have a distance of 4
The area formula for a triangle is 1/2bh and in this case 1/2(1)(4) = 2
The area is 2
