A mixed number always consists of a _____ number plus a fraction.
1 answer:
You might be interested in
First let's talk about the blue line.
You can see its rising so its slope is certainly positive. But by how much is it rising? You can observe that each unit it rises it goes 1 forward and 1 up so its slope is the ratio of 1 up and 1 forward which is just 1.
We have thusly,
Now look at where blue line intercepts y-axis, -1. That is our n.
So the blue line has the equation of,
Next the black lines. The black lines are axes so their equations are a bit different.
First let's deal with x-axis, does it have slope? Yes but it is 0. The x-axis is still, not rising nor falling. Where does x-axis intercept y-axis? At 0. So the equation would be,
Now we have y-axis. Does y axis have a slope? Yes but it is . The y-axis rises infinitely in no run. Where does it intercept y-axis? Everywhere! So what should the equation be? What if we ask where does y-axis intercept x-axis and write its equation in terms of x. Y-axis intercepts x-axis at 0 which means its equation is,
That is, every point of a form lies on y-axis.
Hope this helps :)
Answer:
The equation does not have a real root in the interval
Step-by-step explanation:
We can make use of the intermediate value theorem.
The theorem states that if is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:
- If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
- The image of a continuous function over an interval is itself an interval.
Of course, in our case, we will make use of the first one.
First, we need to proof that our function is continues in , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval
, which means to evaluate the equation in 0 and 1:
Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval . I attached a plot of the equation in the interval
where you can clearly observe how the graph does not cross the x-axis in the interval.
