Miles Traveled Ordered Pair
240 (2, 240)
360 (3, 360)
480 (4,480)
Answer:
15
Step-by-step explanation:
Regina: 6.5h
Phil: 3(2.5h+5)
When you plug in the number of hours, you get the weekly pay. In order to get the same weekly pay, set the equations equal to each other.
6.5h=3(2.5h+5), This is an equation to find the number of hours it will take for them to get equal pay. Now, simplify.
6.5h=7.5h+15
-h=15
h=-15, -15 hours is your answer. You can't work negative hours though which means that Regina and Phil will never be able to make the same amount.
Phil has a higher rate and starting value, this means that Regina will never make the same amount working the same hours.
We need to graph this equation:
Its solutions are the points through which it graph passes. Since it's a linear equation its graph is a straight line and we only need two of its points to draw it. But before graphing let's re-write the equation. We can substract 16x from both sides:
And we divide both sides by 2:
So now with this equation if we pick two random x values we'll get their corresponding y values. This way we'll find two points that are part of the graph which is the line that passes through both. We can begin with x=0:
So the first point is (0,150). Then we can take x=10 and we get:
So the second point is (10,70). Then the graph is the line that passes through points (0,150) and (10,70). In order to represent it
The answer is BFG because those angles form a triangle while the other options form pyramids
Answer:
The following are the answer:
In option a "No".
In option b "Yes".
Step-by-step explanation:
In choice a:
Ax = 0 has no nontrivial solution. A would be the three-pivot matrix, it may assume, that the function has no free variable, and only if the function has had at least one free factor are their nontrivial formulas for the equations of the form Ax=0.
It implies that since A is a 3x3 matrix, has no free variables so that it has no non-trivial choices, and Ax = 0.
In choice b:
we assume that every potential has at least one solution that is Ax=b
. If A does have a three-pivot matrix, It will be a pivot element for each row and column, and for each possible b∈ R³, Ax = b has at least one solution.
