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3 years ago

5

Which graph represents the solution set of the system of inequalities?

2 answers:

Gala2k [10]3 years ago

6 0

the solid line passes through (0,-3) and (0,9)

dotted line passes through (0,-3) and (0,-1)

see attached sketch

Paladinen [302]3 years ago

3 0

Answer:

The image shows the solution for the given inequalities.

Step-by-step explanation:

We are given the following information in the question:

$3y \geq x-9\\3x+y>-3$

We have to plot the solution for the given inequality.

The image attached shows the solution to the given inequalities.

The red shaded portion is the solution to the first inequality and the blue shaded region is the solution for the second inequality.

The common shaded portion is the solution common to both the inequalities.

The dotted blue line is not a part of the solution for both the inequality.

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Write the equation of a possible rational

kondor19780726 [428]

Answer:

The graph has a removable discontinuity at x=-2.5 and asymptoe at x=2, and passes through (6,-3)

Step-by-step explanation:

A rational equation is a equation where

$\frac{p(x)}{q(x)}$

where both are polynomials and q(x) can't equal zero.

1. Discovering asymptotes. We need a asymptote at x=2 so we need a binomial factor of

$(x - 2)$

in our denomiator.

So right now we have

$\frac{p(x)}{(x - 2)}$

2. Removable discontinues. This occurs when we have have the same binomial factor in both the numerator and denomiator.

We can model -2.5 as

$(2x + 5)$

So we have as of right now.

$\frac{(2x + 5)}{(x - 2)(2x + 5)}$

Now let see if this passes throught point (6,-3).

$\frac{(2x + 5)}{(x - 2)(2x + 5)} = y$

$\frac{(17)}{68} = \frac{1}{4}$

So this doesn't pass through -3 so we need another term in the numerator that will make 6,-3 apart of this graph.

If we have a variable r, in the numerator that will make this applicable, we would get

$\frac{(2x + 5)r}{(2x + 5)(x - 2)} = - 3$

Plug in 6 for the x values.

$\frac{17r}{4(17)} = - 3$

$\frac{r}{4} = - 3$

$r = - 12$

So our rational equation will be

$\frac{ - 12(2x + 5)}{(2x + 5)(x - 2)}$

or

$\frac{ - 24x - 60}{2 {x}^{2} + x - 10}$

We can prove this by graphing

5 0

2 years ago

Use the graph to estimate the value of y when x = 2.5<br>

WINSTONCH [101]

Answer:

7 ish

Step-by-step explanation:

if you look at where 2.5 would be on the X axis (in-between 2 and 3) and you look at the corresponding Y point, its somewhere near where 7 would be.

8 0

3 years ago

Need help on this please

KonstantinChe [14]

Answer:

$x^{2}$ + (y-5)² = 8

7 0

2 years ago

The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r

mash [69]

Answer:

$Area = -2.3147$

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates $(2,0)$

The area is represented as:

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

Where

$r = 1 + \cos \theta$

and

$(a,b) = (2,0)$

Substitute values for r, a and b in

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta$

Expand

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta$

By integratin the above, we get:

$Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]$

$Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]$

Substitute 0 and 2 for $\theta$ one after the other

$Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}$

$Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{-sin(2)(cos(2) + 4) - 6}{4}$

Get sin(2) and cos(2) in radians

$Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}$

$Area = \frac{-9.2588}{4}$

$Area = -2.3147$

3 0

2 years ago

List 4 multiples for 1/3

MariettaO [177]

Answer:

2/3 1 4/3 and 5/3

Step-by-step explanation:

just multiply 1/3 with any whole number.

5 0

3 years ago

Read 2 more answers