Question #1
Find the straight line that passes through point (2,4) and is perpendicular to line:
2x+12y-6=0
Rewrite the expression of the line in the following form:
12y=-2x+6y=-16x+12
We know that the slope of the line that is perpendicular to the line with slope k is:
k'=-1k
That’s why the slope of this line for our case is:
k'=-1-16=6
And the equation of the perpendicular line is:
y=6x+b
Where b must be determined knowing that the line passes the point (2, 4). Hence:
4=6*2+bb=4-12=-8
The equation of the straight line that passes through point (2,4) and is perpendicular to line given is:
y=6x-8
Given the function
y=3x2-11x+6=0
What is the value of x?
We have to solve the quadratic equation here:
3x2-11x+6=0D=-112-4*3*6=121-72=49D=7x1,2=11±72*3x1=186=3x2=46=23
Answer:
x1=186=3x2=46=23
Question #2
Given the following information:
Total revenue:
TRX=400*50=20000TRY=500*80=40000
Total cost:
TCX=400*10+13+16+2000=17600TCY=500*12+14+18+1000=23000
Total profit:
TPX=TRX-TCX=20000-17600=2400TPY=TRY-TCY=40000-23000=17000
Breakeven point:
For X:
TPX=0=Qx*50-Qx*10+13+16-2000Qx=200011≈181.82
The breakeven point is at 182 units of X.
TPX=182=2
For Y:
TPY=0=QY*80-QY*12+14+18-1000QY=2509≈27.778
The breakeven point is at 28 units of Y.
TPY=28=8
c.
If profit should be 1000, then:
Qx*50-Qx*10+13+16-2000=1000x=300011≈273 units
QY*80-QY*12+14+18-1000=1000QY=5009≈56 units
Question #3
Explain the differences
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