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2 years ago

Create a budget that will allow you to save at least $100 by the end of October.

Mathematics

1 answer:

insens350 [35]2 years ago

3 0

According to the updated budget, the net income should be at least $2100.

A budget is an estimated strategy for a certain timeframe, mostly one year. Additionally, projected material amounts, expenditures, liabilities, holdings, responsibilities, and working capital may be included.

The following list explains why the figures shown above are accurate:

The factors for the costs in the previous budget are;

Net income =$1,850.00

Cost of rent =$600

Gas, vehicle payment, and insurance together = $475.

Cost of phone, online world, and utility companies = $230.

Cost of grocery stores= $300

Cost of amusement= $50

Cost of homeowners insurance = $20.

$75 has been set aside for recreational expenditures.

Net monthly savings = $100.

Modifications to the spending plan:

$50 is the allotted sum for a computer system.

Monthly net savings = $300

∴Savings increases= $300-$100=$200

Previous net income+ cost of computer system+ 200(extra savings than previous)

= $(1850+50+200)

=$2100

So,Net earnings = $2,100

Learn more about the budget plan:

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To solve this problem you must apply the proccedure below:

1. Let's call:

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2. Then, you must make a system of equations, as below:

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There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression:

$x_v=\frac{-b}{2a}$

Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$

See the graph produced in the attached image.

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