Write an equation for each line in slope-intercept form.

Mathematics

1 answer:

Lady bird [3.3K]4 years ago

3 0

Answer:

y = x + 7

Step-by-step explanation:

The slope-intercept form of a line is:

y = mx + b

where m is the slope and b is the y-intercept.

Looking at the graph, we can see that the line intersects the y-axis at y = 7. So 7 would be our y-intercept.

To find the slope, we would divide the rise of the line by the run. Or m = rise/run. From looking at the graph, we can see that for every 1 unit the line moves in the x-direction, the line moves in the y-direction by 1 unit. Therefore, the rise would be 1 and the run would be 1. 1/1 = 1 so the slope of the line would be 1.

Plugging in 7 for b and 1 for m into the equation for the slope-intercept form, we get:

y = x + 7

So that would be the equation for the line in slope-intercept form.

I hope you find my answer and explanation to be helpful. Happy studying.

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Because of Theorem 5.47 any function that is continuous on (0, 1) but unbounded cannot be uniformly continuous there. Give an ex

almond37 [142]

Answer:

$f: (0,1) \to \mathbb{R}$

$f(x) = \sin(1/x)$

Step-by-step explanation:

f is continuous because is the composition of two continuous functions:

$g(x) = \sin(x)$ (it is continuous in the real numbers)

$h(x) = 1/x$ (it is continuous in the domain (0,1))

It is bounded because $-1 \leq \sin(\theta) \leq 1$

And it is not uniformly continuous because we can take $\varepsilon = 1$ in the definition. Let $\delta > 0$ we will prove that there exist a pair $x,y\in \mathbb{R}$ such that $|x-y|< \delta$ and $|f(x) -f(y)|> \varepsilon = 1$ .