Y=25x+350
y=67x+25x or y=92x
x is the amount of days because y is the total
92x=25x+350 This equation will be how many days the cost will be equal
67x=350
<span>x=5.223</span>
![\bf \begin{array}{clclll} -6&+&6\sqrt{3}\ i\\ \uparrow &&\uparrow \\ a&&b \end{array}\qquad \begin{cases} r=\sqrt{a^2+b^2}\\ \theta =tan^{-1}\left( \frac{b}{a} \right) \end{cases}\qquad r[cos(\theta )+i\ sin(\theta )]\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Bclclll%7D%0A-6%26%2B%266%5Csqrt%7B3%7D%5C%20i%5C%5C%0A%5Cuparrow%20%26%26%5Cuparrow%20%5C%5C%0Aa%26%26b%0A%5Cend%7Barray%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ar%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5C%5C%0A%5Ctheta%20%3Dtan%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7Bb%7D%7Ba%7D%20%5Cright%29%0A%5Cend%7Bcases%7D%5Cqquad%20r%5Bcos%28%5Ctheta%20%29%2Bi%5C%20sin%28%5Ctheta%20%29%5D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
now, notice, there are two valid angles for such a tangent, however, if we look at the complex pair, the "a" is negative and the "b" is positive, that means, "x" is negative and "y" is positive, and that only occurs in the 2nd quadrant, so the angle is in the second quadrant, not on the fourth quadrant.
thus
![\bf 12\left[cos\left( \frac{2\pi }{3} \right) +i\ sin\left( \frac{2\pi }{3} \right) \right]](https://tex.z-dn.net/?f=%5Cbf%2012%5Cleft%5Bcos%5Cleft%28%20%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%5Cright%29%20%2Bi%5C%20sin%5Cleft%28%20%5Cfrac%7B2%5Cpi%20%7D%7B3%7D%20%5Cright%29%20%5Cright%5D)
Answer: -32
You are taking away from "withdrawal"
Answer:
x = 2/3 or x = -1
Step-by-step explanation by completing the square:
Solve for x:
3 x^2 + x - 2 = 0
Divide both sides by 3:
x^2 + x/3 - 2/3 = 0
Add 2/3 to both sides:
x^2 + x/3 = 2/3
Add 1/36 to both sides:
x^2 + x/3 + 1/36 = 25/36
Write the left hand side as a square:
(x + 1/6)^2 = 25/36
Take the square root of both sides:
x + 1/6 = 5/6 or x + 1/6 = -5/6
Subtract 1/6 from both sides:
x = 2/3 or x + 1/6 = -5/6
Subtract 1/6 from both sides:
Answer: x = 2/3 or x = -1
Your question answers itself. You solve by graphing.
I like to subtract one side of the equation from the other, so the solutions are where the graph crosses the x-axis (the resulting function value is zero).
It can be useful to find the "turning point" of each absolute value expression (where its value is zero) and graph that and some points on either side.
