What is the domain and range of the function f(x)= √x
1 answer:
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<span><u><em>Answer:</em></u>
d = 25
<u><em>Explanation:</em></u>
To get the value of d, we will need to isolate the d on one side of the equation.
<u>This can be done as follows:</u>
</span>
<span>
<u>First, we will do cross multiplication as follows:</u>
40*15 = 24*d
600 = 24d
<u>Then, we will isolate the d as follows:</u>
24d = 600
</span>
<span>
d = 25
Hope this helps :)</span>
d = 25
<u><em>Explanation:</em></u>
To get the value of d, we will need to isolate the d on one side of the equation.
<u>This can be done as follows:</u>
</span>
<u>First, we will do cross multiplication as follows:</u>
40*15 = 24*d
600 = 24d
<u>Then, we will isolate the d as follows:</u>
24d = 600
</span>
<span>
d = 25
Hope this helps :)</span>
I: x-2y>=8
II:2x+y<=3
start by isolating y on a side
with I:
x-2y>=8
-2y>=8-x
2y<=x-8
y<=x/2-4
with II:
2x+y<=3
y<=-2x+3
now you can draw I' and II' and read of the solutions with the graph.
transform the inequalities into lines in slope intercept form by replacing the inequality with equality:
y<=x/2-4
->
y=x/2-4
y<=-2x+3
->
y=-2x+3
look at the inequalities to figure out if the area above or below is valid for each line (y<=... -> area below is valid->colored in the attached image)
the area in which all areas overlap is the solution space.
II:2x+y<=3
start by isolating y on a side
with I:
x-2y>=8
-2y>=8-x
2y<=x-8
y<=x/2-4
with II:
2x+y<=3
y<=-2x+3
now you can draw I' and II' and read of the solutions with the graph.
transform the inequalities into lines in slope intercept form by replacing the inequality with equality:
y<=x/2-4
->
y=x/2-4
y<=-2x+3
->
y=-2x+3
look at the inequalities to figure out if the area above or below is valid for each line (y<=... -> area below is valid->colored in the attached image)
the area in which all areas overlap is the solution space.