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Svetradugi [14.3K]
3 years ago
12

Graham graphs the following system of equations to determine its solution.

Mathematics
1 answer:
V125BC [204]3 years ago
3 0

Answer: (2,2)

Step-by-step explanation:

I'm going to use substitution to solve this:

First, I use the second equation to solve for y in terms of x.

2x - y = 2

-y = 2 - 2x

y = 2x - 2

Then, I plug the value for y into the first equation:

2x + 4y = 12

2x + 4(2x - 2) = 12

2x + 8x - 8 = 12

10x - 8 = 12

10x = 20

x = 2

Now that we have x, we can plug it into either equation (I used the first) to get the value for y:

2x + 4y = 12

2(2) + 4y = 12

4 + 4y = 12

4y = 8

y = 2

We get the point (2,2) as a solution.

Note: This is the point where both lines intersect.

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Using the Division algorithm to find q and r such that 3662 = q·16+r , where 0 ≤ r < 16 . What if we take c = −3662 instead o
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Answer:

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b) If c=-3662, then q=-229 and r=2

Step-by-step explanation:

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Then 3662=228*16+14.

b) Observe that -228*16=-3648 and -3648-14=-3662, but r= must be positive. Then -228 doesn't work.

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4 0
3 years ago
If the endpoints of the diameter of a circle are (−6, 6) and (6, −2), what is the standard form equation of the circle?
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(6-6/2}, (-2+6/2)
(0, 2)
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6 0
2 years ago
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Answer:

A

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4 0
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