WHO CAN HELP ME??????

Vesnalui [34]

<h2>Answer:</h2>

Shown below

<h2>Step-by-step explanation:</h2>

No graph has been plotted, but the question is answerable either way and I'll be happy to help you. In this problem, we have the following inequality:

$x-y-2\geq 0$

Before we focus on getting the shaded region, let's graph the equation of the line:

$x-y-2=0$

So let's write this equation in slope intercept form $y=mx+6$ :

STEP 1: Write the original equation.

$x-y-2=0$

STEP 2: Subtract -x from both sides.

$x-y-2-x=0-x \\ \\ \\ Group \ like \ terms \ on \ the \ left \ side: \\ \\ (x-x) - y-2=-x \\ \\ The \ x's \ cancel \ out \ on \ the \ left: \\ \\ -y-2=-x$

STEP 3: Add 2 to both sides.

$-y-2+2=-x+2 \\ \\ -y=-x+2$

STEP 4: Multiply both sides by -1.

$(-1)(-y)=(-1)(-x+2) \\ \\ y=x-2$

So, $m=1$ and $b=-2$ . The graph of this line passes through these points:

$If \ x=0 \ then: \\ \\ y=x-2 \therefore y=(0)-2 \therefore y=-2 \\ \\ Passes \ through \ (0,-2) \\ \\ \\ If \ y=0 \ then: \\ \\ y=x-2 \therefore 0=x-2 \therefore x=2 \\ \\ Passes \ through \ (2,0)$

By plotting this line, we get the line shown in the first figure below. To know whether the shaded region is either above or below the graph, let's take point (0,0) to test this, so from the inequality:

$x-y-2\geq 0 \\ \\ Let \ x=y=0 \\ \\ 0-0-2\geq 0 \\ \\ -2\geq 0 \ False!$

Since this statement is false, then the conclusion is that the region doesn't include the origin, so the shaded region is below the graph as indicated in the second figure below. The inequality includes the symbol ≥ so this means points on the line are included in the region and the line is continuous.

4 0

3 years ago

Please help ! Please help!

son4ous [18]

Answer:

x=139 i believe not 100% about it though its been years since ive done this work.

Step-by-step explanation:

6 0

3 years ago

Last question I need done for my assignment!!! Help would be greatlyyyy appreciated!

sergejj [24]

Answer: $22\ units^2$

Step-by-step explanation:

You need to remember that the area of a rectangle can be found using this formula:

$A=lw$

Where "A" is the area of the rectangle, "l" is the length and "w" is the width.

As you can observe in the picture attached, you can divide the figure into four rectangles (identified as A,B,C and D) whose dimensions are also shown in the picture.

Then, applying the formula, you get that the area of each rectangle is:

$A_A=l_Aw_A=(6\ units)(1\ unit)=6\ units^2\\\\A_B=l_Bw_B=(4\ units)(2\ units)=8\ units^2\\\\A_C=l_Cw_C=(3\ units)(2\ units)=6\ units^2\\\\A_D=l_Dw_D=(2\ units)(1\ unit)=2\ units^2$

Therefore, you need to add the areas of the rectangles calculated before in order to find the area of the figure.

You get that this is:

$A_f=6\ units^2+8\ units^2+6\ units^2+2\ units^2\\\\A_f=22\ units^2$

7 0

3 years ago

Hii please help asap ill give brainliest thanks

cupoosta [38]

Answer:

Buddism

Step-by-step explanation:

The lethal war with Kalinga transformed the vengeful Emperor Ashoka into a stable and peaceful emperor, and he became a patron of Buddhism. According to the prominent Indologist, A. L. Basham, Ashoka's personal religion became Buddhism, if not before, then certainly after the Kalinga War.

5 0

3 years ago

Please I really need help

ohaa [14]

9514 1404 393

Answer:

scale factor: 3
rule: (x, y) ⇒ (3x +15, 3y -24)
center: (-7.5, 12)

Step-by-step explanation:

The scale factor can be found by comparing the length of CB to the length of RQ.

B-C = (-2, 8) -(-4, 9) = (2, -1)

Q-R = (9, 0) -(3, 3) = (6, -3)

The length of RQ is clearly 3 times the length of CB, so the scale factor (k) is ...

ratio of corresponding differences = 6/2 = -3/-1 = 3 . . . . . scale factor

__

We know that for dilation about a point O, the distance from O is multiplied by the scale factor. For dilation of point B to point Q, this means ...

k(B -O) = (Q -O)

Solving for Q, we get ...

Q = kB -kO +O . . . . this is what our dilation rule will look like.

The quantity O-kO can be found by subtracting kB:

Q -kB = O -kO = O(1 -k)

This is what we need for our dilation rule.

Q -kB = (9, 0) -3(-2, 8) = (9+6, 0-24) = (15, -24)

So, our dilation rule is ...

(x, y) ⇒ k(x, y) +(15, -24)

(x, y) ⇒ (3x +15, 3y -24) . . . . . dilation rule

__

The center of dilation can be found from ...

(Q -kB)/(1 -k) = O

O = (15, -24)/(1 -3) = (-7.5, 12)

The center of dilation is (-7.5, 12).

_____

<em>Additional comments</em>

On the graph, the center of dilation can be found by drawing a line through a point and its image. (Points are always dilated along a line through the center of dilation.) The intersection of two such lines is the center of dilation.

On this graph, the center is just above the top edge of the chart, at point (-7.5, 12). You can see this if you carefully draw lines BQ and AP.

You usually have coordinates for two original points and two image points, so finding the scale factor the way we did is not difficult. If you just have one point on the original and the image, the scale factor is found by finding their distances from the center of dilation. Each image point is k times as far as each original point from that center.

7 0

3 years ago