The correct answer is A. Each of the y values are 4 times their respective x values
Answer:
<h2>
No, Given relation is not a function.</h2>
Explanation:
We know that , if any vertical line cuts the given graph of relation at exactly one point, then the relation can be called as function.
From Given graph , we find that the vertical line through any point on x-axis greater than zero (ex : X = 5) cuts the graph at more than one point.
Hence, Given relation is not a function.
Hope this helps...
Good luck on your assignment...
C. "David's equation is correct, because their spending will be multiplied by the number of months and then subtracted from their savings"
they are technically saying "Look we have 12,350 to use. How long (y) will the savings (12,350) last after using 240 each month (x)
so the amount they use *PER* month gets subtracted from the savings because they are using that money to pay with
Hope this helps :)
I think it would be a measuring cup. hope i helped
Answer:
The equation does not have a real root in the interval ![\rm [0,1]](https://tex.z-dn.net/?f=%5Crm%20%5B0%2C1%5D)
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.