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MrRa [10]
2 years ago
9

Lewis has a net spendable income of $2,000 per month. He sets up the following transportation budget for himself.

Mathematics
1 answer:
Yuliya22 [10]2 years ago
7 0

What Lewis did wrong was that; He budgeted more than the maximum recommended amount of money for transportation.

<h3>How to budget effectively?</h3>

We are given;

Net spendable income = $2,000 per month

Now, we see how much he has budgeted for other items such as Car payments, Gas/Oil, Insurance, License/Registration, Taxes, Maintenance/Repair.

Now, from the recommended transportation budget of between 15% - 20% of the  net spendable income, we can see that he would likely spend more than that if we add the cost of maintenance and repair.

Thus, what Lewis did wrong was that he budgeted more than the maximum recommended amount of money for transportation.

Read more about Budget at; brainly.com/question/6663636

#SPJ1

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find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by
coldgirl [10]

Let point P be with coordinates (x_0,y_0). Find the equation of the  tangent line.

1. If y=1-x^2, then y'=-2x.

2. The equation of the tangent line at point P is

y-y_0=-2x_0(x-x_0).

Find x-intercept and y-intercept of this line:

  • when x=0, then y=y_0+2x_0^2;
  • when y=0, then x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.

Since point P is on the parabola, then y_0=1-x_0^2 and

A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.

Find the derivative A':

A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.

Equate this derivative to 0, then

12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.

And

y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.

Answer: two points: P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).

6 0
2 years ago
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