The region is in the first quadrant, and the axis are continuous lines, then x>=0 and y>=0
The region from x=0 to x=1 is below a dashed line that goes through the points:
P1=(0,2)=(x1,y1)→x1=0, y1=2
P2=(1,3)=(x2,y2)→x2=1, y2=3
We can find the equation of this line using the point-slope equation:
y-y1=m(x-x1)
m=(y2-y1)/(x2-x1)
m=(3-2)/(1-0)
m=1/1
m=1
y-2=1(x-0)
y-2=1(x)
y-2=x
y-2+2=x+2
y=x+2
The region is below this line, and the line is dashed, then the region from x=0 to x=1 is:
y<x+2 (Options A or B)
The region from x=2 to x=4 is below the line that goes through the points:
P2=(1,3)=(x2,y2)→x2=1, y2=3
P3=(4,0)=(x3,y3)→x3=4, y3=0
We can find the equation of this line using the point-slope equation:
y-y3=m(x-x3)
m=(y3-y2)/(x3-x2)
m=(0-3)/(4-1)
m=(-3)/3
m=-1
y-0=-1(x-4)
y=-x+4
The region is below this line, and the line is continuos, then the region from x=1 to x=4 is:
y<=-x+2 (Option B)
Answer: The system of inequalities would produce the region indicated on the graph is Option B
The answer is C. hope this helps.
The answer is the third option.
The solution is in the pdf file attached. Please, let me know if you can follow the steps.
Hello from MrBillDoesMath!
Answer: (1/7) * ( 4 +\- sqrt(5) i)
where i = sqrt(-1)
Discussion:
The solutions of the quadratic equation ax^2 + bx + c = 0 are given by
x = ( -b +\- sqrt(b^2 - 4ac) )/2a.
The equation 7 x^2 + 3 = 8x can be rewritten as
7x^2 - 8x + 3 = 0.
Using a = 7, b = -8 and c = 3 in the quadratic formula gives:
x = (8 +\- sqrt ( (-8)^2 - 4*7*3) ) / (2*7)
= ( 8 +\- sqrt( 64 - 84)) /(2*7)
= ( 8 +\- sqrt( -20) ) / (2*7)
= ( 8 +\- sqrt( -20) ) / 14
= 8/14 +\- sqrt(5 *4 * -1) /14
= 4/7 +\- 2 sqrt(5) *i /14
As 2/14 = 1/7 in the second term
= 4/7 +\- sqrt(5) *i /7
Factor 1/7 from each term.
= (1/7) * ( 4 +\- sqrt(5) i)
Thank you,
MrB
Answer: b. 60 movies
Step-by-step explanation:
60 * $5= $300
$300+$120= $420
$420/60= $7
