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ZanzabumX [31]
3 years ago
12

Using a table to graph equations in standard form help.Be dumb i will report ​

Mathematics
1 answer:
sertanlavr [38]3 years ago
8 0

Answer:

X, Process, Y

-2, y = -2 - 3, -5

-1, y = -1 - 3, -4

0, y = 0 - 3, -3

1, y = 1 - 3, -2

2, y = 2 - 3, -1

X = -2, -1, 0, 1, 2

Y = -5, -4, -3, -2, -1

Step-by-step explanation:

Coordinates to the graph are (-2,-5) (-1,-4) (0,-3) (1,-2) (2,-1)

Graph is the image below.

you want to get Y by itself so here is how you do that.

-x+y=-3\\+x       +x\\

Now we have y=x-3

now we plug x into the equation to get our answer.

-2 - 3 = -5

-1 - 3 = -4

0 - 3 = -3

1 - 3 = -2

2 - 3 = -1

Hope this helps!

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Which ordered pair can be plotted together with these four points, so that the resulting graph still represents a function?
MrRissso [65]

Answer:

B.(-1,2)

Step-by-step explanation:

In a function, there can not be two different values of y corresponding to the same value of x.

See the graph attached.

Here, the points on the graph are (1,2), (2,-3), (-2,-2) and (-3,1).

If we consider point (-2,2) then there will be two points corresponding to the same x value i.e. (-2,-2) and (-2,2).

Similarly, if we consider the point (2,-2) or (2.-1) then also there becomes more than one values of y for a single value of x i.e. x = 2.

So, if we consider the ordered pair (-1,2) then only the graph still represents a function. (Answer)

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3 years ago
Ravi walks 3.5km every day. How far does he walk in 7 days?
kolbaska11 [484]
24.5 you take 3.5 and times it by 7 do get the distance.

4 0
3 years ago
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Three six-sided fair dice are rolled. The six sides of each die are numbered 1; 2; : : : ; 6. Let A be the event that the first
valkas [14]

Answer:

1.) [A, B, C]

2.) [A', B', C']

3.) [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

4.) A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

Step-by-step explanation:

If A is the event that the first die shows an even number, then A' is the event that first die DOES NOT show an even number.

If B is the event that the Second die shows and even number, then B' is the event that second die DOES NOT show an even number.

If C is the event that the third die shows an even number, then C' is the event that third die DOES NOT show an even number.

Hence, our possible events are A, A', B, B', C, C'.

For question 1, in the event that all three dice show and even number, the expression of the event = [A, B, C]

For question two, in the event that no die shows an even number, the expression of the event = [A', B', C']

For question 3,In the event that at least one die shows an odd number, the expression of the event = [A', B, C] or [A, B', C] or [A, B, C'] or [ A', B', C] or [A', B, C'] or [A, B', C'] or [A', B', C']

For question 4,in the event that at most two dice show odd numbers, the expression of the event = [A', B', C] or [A', B, C'] or [A, B', C'] or [A', B, C] or [A, B', C] or [A, B, C'] or [A, B, C]

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3 years ago
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Oduvanchick [21]

Answer:

(- 1, 1 )

Step-by-step explanation:

Given the 2 equations

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