.. pls help me....!!!!! Determine whether the point (-2,4) satisfy y<3x+5,y=3x+5 or y >3x+5..

Mathematics

1 answer:

natima [27]3 years ago

3 0

Given:

The point is (-2,4).

To find:

Whether the point (-2,4) satisfy $y$ or $y >3x+5$ .

Solution:

We have a point (-2,4).

Substituting x=-2 in 3x+5 and check whether it is greater, equal or less than y value, i.e., 4.

$3x+5=3(-2)+5$

$=-6+5$

$=-1$

-1 is less than 4.

$3x+5$

In can be written as

$y>3x+5$

Therefore, the given point (-2,4) satisfy $y >3x+5$ .

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Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.