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Murrr4er [49]
2 years ago
14

Solve the following system of equations graphically on the set of axes below.

Mathematics
1 answer:
eimsori [14]2 years ago
8 0

Answer:

  (x, y) = (3, 5)

Step-by-step explanation:

To solve these equations graphically, you plot each one, then identify the coordinates of the point of intersection.

The first equation can be graphed using the y-intercept of -1 and the slope of 2.

The second equation has a y-intercept of 18/3 = 6, and a slope of -1/3. The x-intercept of 18 is off the chart, so the slope and intercept are a good way to plot this line, too.

__

The slope is the "rise" divided by the "run". A slope of 2 means the line goes up 2 units for each unit to the right. A slope of -1/3 means the line goes down 1 unit for each 3 units to the right.

The solution is (x, y) = (3, 5).

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If a(x) = 3x + 1 and b(x)= squareroot x-4, what is the domain of (b*a)(x)?
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\boxed{ \ x \geq 1 \ } or can be written as \boxed{ \ [1, \infty) \ }

<h3>Further explanation</h3>

This is a question about the composition of functions and how to get a domain function.

Given \boxed{ \ a(x) = 3x + 1 \ } and \boxed{ \ b(x) = \sqrt{x - 4} \ }.

We will form (b o a)(x) and then determine the domain.

<u>Step-1</u>

\boxed{ \ (b \circ a)(x) = b(a(x)) \ }

Replace each appearance of x in b(x) with \boxed{ \ a(x) = 3x + 1 \ }.

\boxed{ \ (b \circ a)(x) = \sqrt{(3x + 1) - 4} \ }

Thus, \boxed{ \ (b \circ a)(x) = \sqrt{3x - 3} \ }

<u>Step-2</u>

To be defined, the value under the radical sign must not be negative. Therefore, the domain of (b \circ a)(x) = \sqrt{3x - 3} are processed as follows.

\boxed{ \ 3x - 3 \geq 0 \ }

Both sides added by 3.

\boxed{ \ 3x \geq 3 \ }

Both sides divided by 3.

\boxed{ \ x\geq 1 \ }

Thus, the domain of (b \circ a)(x) = \sqrt{3x - 3} is \boxed{ \ x \geq 1 \ } or can be written as \boxed{ \ [1, \infty) \ }

<h3>Learn more</h3>
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4 years ago
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