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ser-zykov [4K]
3 years ago
15

The system of equations shown below is graphed on a coordinate grid:

Mathematics
1 answer:
musickatia [10]3 years ago
5 0

First of all, for a point to be the solution to the system of equations, it must be located on both lines, since we are calculating the point at which the lines cross. Therefor, we can automatically discount B and D as the answers, which state that the point does not lie on either line. To see whether it is A or C we will have to solve the equations.

1. There are quite a few ways to approach solving the system of equations, however I will show the substitution method as it requires less modification of equations.

What we need to do is to isolate either x or y in one of the equations so that we can then substitute this value into the other equation. So for example if we take the first equation and isolate x:

3y + x = 6

x = 6 - 3y (Subtract 6 from both sides)

2. Now we can substitute this into the second equation (2y - x = 9).

when x = 6 - 3y:

2y - (6 - 3y) = 9

2y - 6 + 3y = 9

5y - 6 = 9 (Add 2y and 3y)

5y = 15 (Add 6 to both sides)

y = 3 (Divide both sides by 5)

From here, we can already see that the answer is C, since A has a y-value of 4, however if we were to completely solve the question we would do the following:

3. Now that we know the value of y, we can substitute this back into our equation of x = 6 - 3y to find x:

x = 6 - 3(3)

x = 6 - 9

x = -3

So, our solution is the point (-3, 3). It also has to lie on both lines to be a solution. Therefor C. It is (-3, 3) and lies on both lines, is the correct answer.

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Answer: Option 'B' is correct.

Step-by-step explanation:

Let the initial concentration be

N_0=900\ g

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t_{\frac{1}{2}}=15

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3 years ago
If the point (a, b) is equidistant from the points (-a, 2) and (2, -b), then prove that 3(a+b)+4=0​
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Answer:  see proof below

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

Since (a, b) is equidistant from (-a, 2) and (2, -b), then it is the midpoint of the those two points. Use Midpoint formula to find (a, b).

M_x=\dfrac{x_1+x_2}{2}\qquad \qquad \qquad M_y=\dfrac{y_1+y_2}{2}\\\\\\a=\dfrac{-a+2}{2}\qquad \qquad \qquad \quad b=\dfrac{2-b}{2}\\\\\\2a=-a+2\qquad \qquad \qquad \quad 2b=2-b\\\\\\3a=2\qquad \qquad \qquad \qquad \qquad 3b=2\\\\\\a=\dfrac{2}{3}\qquad \qquad \qquad \qquad \qquad b=\dfrac{2}{3}

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3\bigg(\dfrac{2}{3}+\dfrac{2}{3}\bigg)-4=0\\\\\\3\bigg(\dfrac{4}{3}\bigg)-4=0\\\\\\4-4=0\\\\0=0\qquad \text{TRUE!}

<em>Notice that I changed the equation to "negative 4" because the equation you provided did not make a true statement.</em>

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