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pantera1 [17]
3 years ago
5

Find the slope of the line.

Mathematics
1 answer:
Damm [24]3 years ago
8 0

y2-y1 / x2-x1 is how to get slop3 for 2 points (x1,y1) and (x2, y2). So 4-1 / 2-0 and the slope is 3/2. If this is right could you possibly give me brainliest? Hope this helped.

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Step-by-step explanation:made the mona lisa

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3 years ago
Whice set of side will make a triangle
telo118 [61]

First things first.. What type of triangle

If u are taking about a normal triangle here is some:

∠60+∠60+∠60

∠30+∠90+∠60

∠11+∠82+∠87

7 0
3 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
4x-10y=-16 and -7x+20y=23 elimination
mixas84 [53]

(-5/8, 27/20) is the answer.

6 0
3 years ago
In this exercise, we estimate the rate at which the total personal income is rising in a metropolitan area. In 1999, the populat
DiKsa [7]

Answer:

the rate at which total personal income was rising in the area in 1999 is $1,580,507,200 billion

Step-by-step explanation:

From the given information:

Let consider y to represent the number of years after 1999

Then the population in time (y) can be expressed as:

P(y) = 9400y + 924900

The average annual income can be written as:

A(y) = 1400y + 30388

The total personal income = P(y)  ×  A(y)

The rate at which the total personal income is rising is T'(y) :

T'(y) = P'(y)  ×  A(y)  + P(y)  ×  A'(y)

T'(y) = (9400y + 924900)' (1400y + 30388) + (9400y + 924900) (1400y + 30388)'

T'(y) = 9400(1400y + 30388) + (9400y + 924900) 1400

Since in 1999 y =0

Then:

T'(0) = 9400(1400(0) + 30388) + (9400(0) + 924900) 1400

T'(0) = 9400(30388) + (924900)1400

T'(0) = $1,580,507,200 billion

Therefore; the rate at which total personal income was rising in the area in 1999 is $1,580,507,200 billion

8 0
3 years ago
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