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What are the steps involved in the financial planning process?

1 answer:

rjkz [21]3 years ago

3 0

Answer:

The steps:

Establishing your current financial state
Collecting the desired goals
Analysis of any alternatives
Develop the alternatives
Implement the action plan
Review the plan

Step-by-step explanation:

The steps involved in financial planing are;

Knowing your current financial state
Coming up with financial goals
Determining alternative courses of action
Evaluating the alternatives
Creating and implementing the action plan
Reevaluating and revising the plan

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I need help 2(4+2x)≥5x+5

timofeeve [1]

Answer:

x ≤ 3

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Distributive Property

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

Terms/Coefficients

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

2(4 + 2x) ≥ 5x + 5

<u>Step 2: Solve for </u><em><u>x</u></em>

[Distributive Property] Distribute 2: 8 + 4x ≥ 5x + 5
[Subtraction Property of Equality] Subtract 5x on both sides: 8 - x ≥ 5
[Subtraction Property of Equality] Subtract 8 on both sides: -x ≥ -3
[Division Property of Equality] Divide -1 on both sides: x ≤ 3

5 0

2 years ago

What is the greatest conman factor of -32x3+41x4y

KiRa [710]

Answer:

x^3

Step-by-step explanation:

6 0

3 years ago

Two children own two-way radios that have a maximum range of 3 miles. One leaves a certain point at 1:00 P.M., walking due north

Grace [21]

Answer:

The answer is 21 minutes

Step-by-step explanation:

We use the equation Xf = Xo + vt

1) At 1:00 PM, child one leaves the starting point heading north at a constant velocity of 6 mi/hr or .1 [mi/min] (divide by 60 to convert from [mi/hr] to [mi/min])

2) He walks for 15 minutes before kid 2 starts walking. In 15 minutes he is able to cover 1.5 [mi]

$x_{1f1} =x_{o} +v_{1} t_{1} \\x_{1f1} =0+.1(15)\\x_{1f1} =1.5 [mi]$

3) Now, child 2 starts walking and we know that when the range reaches 3 miles, they won´t be able to communicate. So the sum of the final position of child 1 and child 2 must be 3[mi]

Child 1 final position => $x_{1f} = x_{1f1} +v_{1} t\\x_{1f} =1.5+v_{1} t$
Child 2 final position => $x_{2f} =0+v_{2} t$

4) Sum the equations and equate to 3

$x_{1f} +x_{2f} =3$

5) Substitute the values we already know

$1.5+v_{1} t+v_{2}t=3\\ 1.5+.1t+.15t=3\\1.5+.25t=3\\t=\frac{3-1.5}{.25} \\t=6 [min]$

6) in 15 + 6 minutes they will be 3miles apart

7) In 21 minutes they will still be able to communicate with one another.

7 0

3 years ago

Five times a number is the same as 30 more than 8 times the number. Find the number.

allsm [11]

5x = 8x+30

5x-8x+30 = 0

-3x=-30

x=-10

5 0

3 years ago

Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

<em>Solving the expression</em>

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

<em>Solving the expression</em>

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

<em>Solving the expression</em>

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

As

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

<em>Solving the expression</em>

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

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5 0

3 years ago

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