Find the coordinates of the midpoint for the pair of points below a) (2,5) , (8,9) b) (9,1) , (5,17)

???? Help please you get 40 points if u help

IrinaVladis [17]

Answer:

B) P = 8x + 4

Step-by-step explanation:

Formula:

P= 2( l + w)

P = 2( x + 2 + 3x)

P = 2(4x + 2)

P = 8x + 4

6 0

3 years ago

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Give at least three examples of how earlier periods made use of the accounting process? How did the accounting process help such

Ksju [112]

Answer:

i am not sure of these answer but what i know i am gonnna share eith you

Step-by-step explanation:

. The people relied on primitive accounting methods to record the growth of crops and herds. Because there was a natural season to farming and herding, it was easy to count and determine if a surplus had been gained after the crops had been harvested or the young animals weaned.

2. The use of the words "debit" and "credit" in accounting goes back to the days of single-entry bookkeeping, which had as its chief objective keeping track of amounts owed by customers (debtors) and amounts owed to creditors.

3. Accounting as an ad-hoc ordering system devised by the merchant. Its regular use provides the merchant with continued information about business, and allows to evaluate how things are going and to act accordingly.

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4 0

2 years ago

Need help in number 12 and 13 PLEASEE!! I don’t get it and I start school the day after tomorrow

Natasha_Volkova [10]

Answer:

The explanations for the graphs are provided down below. Please let me know if you have any questions about my answer.

12 and 13 as written on the worksheet is right.

Step-by-step explanation:

12) The answer given is correct.

The relation between x and y is given as:

$y=\frac{x^2}{2}-3$ with $x \in \{-4,-2,0,2}$ .

I replaced the word domain with x since the domain is the set of x's for which the relation exists.

We are going to replace x with each of the x's given to see what y corresponds to each.

Let's begin with x=-4.

$y=\frac{x^2}{2}-3$ with $x=-4$ :

$y=\frac{(-4)^2}{2}-3$

$y=\frac{16}{2}-3$

$y=8-3$

$y=5$ .

So (-4,5) is an ordered pair that should be on our graph.

To find this point you move left 4 from origin then up 5. Now you put a dot where you have landed. Your graph does show this point.

Moving on.

Let's do the next x: x=-2.

$y=\frac{x^2}{2}-3$ with $x=-2$ :

$y=\frac{(-2)^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (-2,-1) is an ordered pair that should be on our graph.

To find this point you move left 2 from origin and then down 1. Now you put a dot where you have landed. Your graph shows this point as well.

Now x=0.

$y=\frac{x^2}{2}-3$ with $x=0$ :

$y=\frac{0^2}{2}-3$

$y=\frac{0}{2}-3$

$y=0-3$

$y=-3$

So (0,-3) is an ordered pair that should be on our graph.

To find this point you move left and right none and down 3. Now you put a dot where you have landed. Your graph shows this point.

Now the last point will be at x=2.

$y=\frac{x^2}{2}-3$ with $x=2$

$y=\frac{2^2}{2}-3$

$y=\frac{4}{2}-3$

$y=2-3$

$y=-1$ .

So (2,-1) is an ordered pair that should be on our graph.

To find this point you move 2 units right from the origin and then down 1 unit. Now put a dot where you landed. The graph shows this point as well.

13) The answer given is correct.

$g(x)=|x|$ is the parent function and makes like a V shaped graph where it's vertex is at (0,0).

If we want to move this graph right 3 it becomes:

$m(x)=|x-3| \text{ or } m(x)=|(-1)(-x+3)|=|-1||-x+3|=1|-x+3|=|-x+3|=|3-x|$ .

If you move that up once it becomes:

$n(x)=|x-3|+1$ or $n(x)=|3-x|+1$ which is the curve given.

If you don't know about transformations you can choose a few points to plug in to see what's going on with the graph.

Let's choose x=-5,-3,-1,0,1,3,5.

x=-5

f(-5)=|3--5|+1=|3+5|+1=|8|+1=8+1=9.

There is no room for (-5,9) on our graph but if you extended the left hand side of the absolute value function there you would see that (-5,9) is reached.

x=-3

f(-3)=|3--3|+1=|3+3|+1=|6|+1=6+1=7.

(-3,7) should be a point on the graph. Same thing for this point as (-5,9).

x=-1

f(-1)=|3--1|+1=|3+1|+1=|4|+1=4+1=5.

(-1,5) is located on the graph.

x=0

f(0)=|3-0|+1=|3|+1=3+1=4.

(0,4) is also located on the graph.

x=1

f(1)=|3-1|+1=|2|+1=2+1=3.

(1,3) is located on the graph.

x=3

f(3)=|3-3|+1=|0|+1=0+1=1.

(3,1) is located on the graph.

x=5

f(5)=|3-5|+1=|-2|+1=2+1=3.

(5,3) is located on the graph.

Now if we weren't given the graph already:

I would plot the points I found which were:

(-5,9)

(-3,7)

(-1,5)

(0,4)

(1,3)

(3,1)

(5,3)

We should get a basic idea of what the function looks like from these points.

I will graph them. You will have to connect these points because the domain isn't discrete like number 12 is. That is they didn't list out elements for the domain.

I'm going to graph one more point after x=5.

How about x=7?

$f(7)=|3-7|+1=|-4|+1=4+1=5$

So (7,5) is also a point on the graph.

You should see that the blue points are following the red path I made there.

8 0

2 years ago

On a coordinate plane, 3 triangles are shown. Triangle B C D has points (1, 4), (1, 2), (5, 3). Triangle B prime C prime D prime

Tresset [83]

The rule which describe the composition of transformations that

maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis composition translation of 6 units x,

negative 5 units y ⇒ last answer

Step-by-step explanation:

Let us revise the reflection across the y-axis , horizontal translation

and vertical translation

1. If point (x , y) is reflected across the y-axis, then its image is (-x , y)

2. If point (x , y) is translated h units to the right, then its image is

(x + h , y), if translated h units to the left, then its image is (x - h , y)

3. If point (x , y) is translated k units up, then its image is (x , y + k),

if translated k units down, then its image is (x , y - k)

∵ The vertices of triangle BCD are (1 , 4) , (1 , 2) , (5 , 3)

∵ The vertices of triangle B'C'D' are (-1 , 4) , (-1 , 2) , (-5 , 3)

∵ The x-coordinates of the vertices of Δ B'C'D' have the same

magnitude of x-coordinates of Δ BCD and opposite signs

∴ Δ B'C'D' is the image of Δ ABC after reflection across the y-axis

∵ The vertices of triangle B'C'D' are (-1 , 4) , (-1 , 2) , (-5 , 3)

∵ The vertices of triangle B''C''D'' are (5 , -1) , (5 , -3) , (1 , -2)

∵ The image of -1 is 5 and the image of -5 is 1

∴ The x-coordinates of the vertices of triangle B'C'D' are added by 6

∵ The image of 4 is -1 , image of 2 is -3 and the image of 3 is -2

∴ The y-coordinates of the vertices of triangle B'C'D' are subtracted

by 5

∴ Δ B"C"D" is the image of Δ B'C'D' by translate 6 units to the right

and 5 units down ⇒ (x + 6 , y - 5)

The rule which describe the composition of transformations that

maps ΔBCD to ΔB"C"D" is:

Reflection across the y-axis composition translation of 6 units x,

negative 5 units y

Learn more:

You can learn more about reflection in brainly.com/question/11203617

#LearnwithBrainly

3 0

2 years ago

Read 2 more answers

Kayla has a tshirt store. she sold three times as many white tshirts as blue tshirts. she sold a total of 48 tshirts. How many w

SpyIntel [72]

First make the equation:

1. x
2. 3x

x + 3x = 48

Now solve it:
x + 3x = 48
4x= 48
x=12

Now plug in x
1. x = 12
2. 3x = 36

So Kayla sold 12 blue t-shirts and 36 white t-shirts.

6 0

3 years ago

Find the coordinates of the midpoint for the pair of points below a) (2,5) , (8,9) b) (9,1) , (5,17)​

Find the coordinates of the midpoint for the pair of points below a) (2,5) , (8,9) b) (9,1) , (5,17)