Answer:
Neither parallel nor perpendicular
Step-by-step explanation:
I'm assuming you meant line k is y = 3x -2. If not, this is wrong.
For this, you need to put both lines in point-slope form, or the form that line k is already in. This means you only need to convert line m.
-2r + 6v = 18
6v = 2r + 18
v = 2/6r + 18/6
v = 1/3r + 3
Now you can answer the question.
To be parallel, lines must have the same slope (but a different y-intercept). 3 and 1/3 are not the same, so the lines are not parallel.
To be perpendicular, one line must have the opposite reciprocal (fraction flipped and + goes to - or - to +) of the other. While 3 is the reciprocal of 1/3, they are both positive, so they are not perpendicular.
To be the same line, the equations must be absolutely identical, which they aren't.
This leaves the last option: neither.
Let me know if you need a more in-depth explanation of anything here! I'm happy to help!
Answer:
Step-by-step explanation:
The original price of the jacket at the store is $50.
Evan has a coupon for 15$ off. This means that the amount that Evans will pay is the original price - 15% of the original price. It becomes
50 - (15/100 × 50) = 50 - 7.5 = $42.5
Max pays only 0.6 of the price because his father works at the store.
This means that the amount that Max will pay is 0.6 × the original price. It becomes
0.6 × 50 = $30
Max will pay the least amount because his amount is smaller
Don’t let the letter scare you, imagine this as a simple cross product!
(32 × 1) ÷ 8 = m
32 ÷ 8 = m
4 = m
There you go, the solution to this equation is that m = 4!
I really hope this helped, if there’s anything just let me know! ☻
The <em>correct answers</em> are:
C) No: we would need to know if the vertex is a minimum or a maximum; and
C)( 0.25, 5.875).
Explanation:
The domain of any quadratic function is all real numbers.
The range, however, would depend on whether the quadratic was open upward or downward. If the vertex is a maximum, then the quadratic opens down and the range is all values of y less than or equal to the y-coordinate of the vertex.
If the vertex is a minimum, then the quadratic opens up and the range is all values of y greater than or equal to the y-coordinate of the vertex.
There is no way to identify from the coordinates of the vertex whether it is a maximum or a minimum, so we cannot tell what the range is.
The graph of the quadratic function is shown in the attachment. Tracing it, the vertex is at approximately (0.25, 0.5875).