Answer:
x=2.125
y=0
C=19.125
Step-by-step explanation:
To solve this problem we can use a graphical method, we start first noticing the restrictions 
and 
, which restricts the solution to be in the positive quadrant. Then we plot the first restriction 
shown in purple, then we can plot the second one 
shown in the second plot in green.
The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.
So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.
Step-by-step explanation:
definition of the derivative to differentiate functions. This tutorial is well understood if used with the difference quotient .
The derivative f ' of function f is defined ascthe above pic.
when this limit exists. Hence, to find the derivative from its definition, we need to find the limit of the difference quotient.
Y4 - y3 + 2y2 + y - 1 over (y + 1) • (y2 - y + 1)
<span>
</span>
Answer:
abcdefghigklmanp der gan gang
Step-by-step explanation:
Hello
<span>an equation for the line in point-slope form and general form is :
y = ax+b a : </span>slope ; the <span>Passing through (x' ; y')
</span>y' = ax'+b
y-y' =a(x-x') and : x' =8 y' = -4
calculate a :
let : y = ax+b .....(D)
....(D') y = (1/5)x+4
.(D) perpendicular to(D') : slope (D) × slope (D') = -1
slope (D') = 1/5
slop(D)×(1/5) = -1
slope (D) = -5
equation for the line : y-y' =a(x-x')
y+4 =(-5) (x-8)
