Which ordered pairs are solutions to the inequality 2x-y>1?

Mathematics

1 answer:

RideAnS [48]3 years ago

8 0

Answer:

Any points in the shaded region including (2,-2) and (-3,-8)

Step-by-step explanation:

Convert the line into slope intercept form and graph it.

2x-y > 1 becomes -y>1-2x. Divide both sides by -1 and you get y<2x-1. Graph it with the shaded area on the right and a dashed line.

Any point which falls within the shaded red of the graph is a solution. No points on the line since it is not equal to (its dashed) are solutions. Check the location of your points to verify that they fall within this area.

(-3, -8) ---Yes

(-1, -3) ---No

(0, 5) --- No

(1, 6) --- No

(2, -2) ---Yes

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Find the perimeter of a field that has length 2/x + 1 and width 5/x^2 -1.

Semmy [17]

Answer:

D)4x + 6/(x + 1)(x - 1)

Step-by-step explanation:

A field is basically a rectangle, so to find the perimeter of our field we are using the formula for the perimeter of a rectangle

$p=2(l+w)$

where

$p$ is the perimeter

$l$ is the length

$w$ is the width

We know from our problem that the field has length 2/x + 1 and width 5/x^2 -1, so $l=\frac{2}{x+1}$ and $w=\frac{5}{x^2-1}$ .

Replacing values:

$p=2(l+w)$

$p=2(\frac{2}{x+1} +\frac{5}{x^2-1})$

Notice that the denominator of the second fraction is a difference of squares, so we can factor it using the formula $a^2-b^2=(a+b)(a-b)$ where $a$ is the first term and $b$ is the second term. We can infer that $a=x^2$ and $b=1^2$ . So, $x^2-1=(x+1)(x-1)$ . Replacing that: