The first line in a system of linear equations has a slope of 3 and passes through the point (-1 , -8). The second line passes t
hrough the points (-6 , 3) and (10 , -1).
What is the solution to the system?
What is the solution to the system?
1 answer:
The equation of the first line can be written in point-slope form as
.. y = 3(x +1) -8
or
.. 3x -y = 5
The equation of the second line can be written in 2-point form as
.. y = (-1-3)/(10-(-6))*(x +6) +3
.. y = (-1/4)(x +6) +3
or
.. x +4y = 6
A graph shows the solution to this system is (x, y) = (2, 1).
_____
The second equation can be used to write an expression for x:
.. x = 6 -4y
This can be substituted into the first equation.
.. 3(6 -4y) -y = 5
.. 18 -13y = 5 . . . . . . . collect terms
.. 13 = 13y . . . . . . . . . add 13y-5
.. 1 = y . . . . . . . . . . . . divide by 13
From the above equation for x
.. x = 6 -4*1 = 2
.. y = 3(x +1) -8
or
.. 3x -y = 5
The equation of the second line can be written in 2-point form as
.. y = (-1-3)/(10-(-6))*(x +6) +3
.. y = (-1/4)(x +6) +3
or
.. x +4y = 6
A graph shows the solution to this system is (x, y) = (2, 1).
_____
The second equation can be used to write an expression for x:
.. x = 6 -4y
This can be substituted into the first equation.
.. 3(6 -4y) -y = 5
.. 18 -13y = 5 . . . . . . . collect terms
.. 13 = 13y . . . . . . . . . add 13y-5
.. 1 = y . . . . . . . . . . . . divide by 13
From the above equation for x
.. x = 6 -4*1 = 2
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Answer:
A. 4.5
Step-by-step explanation:
According to my graphing calculator, e^1.5=4.5 which is rounded to the nearest tenth.
Answer: e^1.5 = 4.482
Step-by-step explanation:
Given:
We need to find the value of function at x=1.5
Put x=1.5 into f(x)
Now we will verify the result using graph.
First we draw the graph of f(x) and a vertical line x=1.5
Please see the attachment for graph.
Hence, The value of
3x -y ⩾ 6
3x - 6 ⩾ y
now, with inequalities, what we do is, we graph the line of 3x - 6 = y, and then we shade the "true region".
if we pick a point on say hmmm (4, 0), namely x = 4 and y = 0, we can plug that in the inequality and see what we get,
3(4) - 0 ⩾ 6
12 - 0 ⩾ 6
12 ⩾ 6
is 12 really greater or equals to 6? well yes, therefore, the point (4, 0) lies on the "true region", since it's true, 12 is indeed ⩾ 6, so, where that point is, we shade.
now, the ⩾ means equals to or greater, and therefore, since the values could also equal the boundary points, the line is a solid line, because it includes the line itself, as well as the shading.
check the picture below.
3x - 6 ⩾ y
now, with inequalities, what we do is, we graph the line of 3x - 6 = y, and then we shade the "true region".
if we pick a point on say hmmm (4, 0), namely x = 4 and y = 0, we can plug that in the inequality and see what we get,
3(4) - 0 ⩾ 6
12 - 0 ⩾ 6
12 ⩾ 6
is 12 really greater or equals to 6? well yes, therefore, the point (4, 0) lies on the "true region", since it's true, 12 is indeed ⩾ 6, so, where that point is, we shade.
now, the ⩾ means equals to or greater, and therefore, since the values could also equal the boundary points, the line is a solid line, because it includes the line itself, as well as the shading.
check the picture below.