Middle School teaches 6th, 7th, and 8th grade classes. The sixth grade has 318 students, the 7th has 286 students, and the 8th h
1 answer:
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Step-by-step explanation:
The domain and range of a function and its inverse are inverted i.e. The domain of the function will become range of its inverse and the range of the function will become the domain of its inverse.
When representing the function in ordered pairs, the first values in the pair represents the domain and second values represent the range of the function.
h(x) = {3,5),(5,-7)(6,-9)(10,-12)(12,16)}
Domain of h(x) is: {3, 5, 6, 10, 12}
Range of h(x) is: (5, -7, -9, -12, 16}
So,
Domain of = Range of h(x) = (5, -7, -9, -12, 16}
Range of = Domain of h(x) = {3, 5, 6, 10, 12}
Therefore, will be represented as:
{(5,3), (-7,5), (-9, 6), (-12, 10), (16,12)}
It seems like you have made some error while writing the original function because of which we are not getting an exact match in the answer. However by looking at the answers, option B is the nearest most and seems the correct answer.
Answer:
A system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2
Step-by-step explanation:
The equation of the given line is 8·x - 4·y = 12
Which gives;
8·x- 12= 4·y
y = 2·x - 3
Given that the line passes through the points (0, -3) and (1, -1), we have;
When (x₁, y₁) = (0. -3) and (x₂, y₂) = (1, -1), we have;
y - (-3) = 2×(x - 0)
y = 2·x - 3 which is the equation of the given line
For the lines 8·x - 4·y = 12, which is the sane as y = 2·x - 3 and the line y = m·x - 6 to have no solution, the slope of the two lines should be equal that is m = 2
Given that the line passes through the point (1.5, 0), we have;
y - 0 = 2×(x - 1.5)
y = 2·x - 3...................(1)
For the equation, y = m·x - 6, when m = 2, we have;
y = 2·x - 6..................(2)
Solving equations (1) and (2) gives;
2·x - 3 = 2·x - 6, which gives;
2·x - 2·x= - 3 - 6
0 = 9
Therefore, a system of parallel lines will be created where the two lines will never meet and have no common solution at a value of m = 2.