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3 years ago

Employee earns $66,000 a year. How much is their bi-weekly gross pay? PLEASE

Mathematics

1 answer:

Natalija [7]3 years ago

7 0

Answer:

Calculate gross pay, before taxes, based on hours worked and rate of pay per hour including overtime. To enter your time card times for a payroll related calculation use this time card calculator. Gross Pay or Salary: Gross pay is the total amount of money you get before taxes or other deductions are subtracted from your salary.

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NEED HELP ASAP What is true about the points on the graph of Y=3^x?

galben [10]

Answer:

Step-by-step explanation:

The answe of this question is d

3 0

3 years ago

4 more than one third of a number n is 6. An equation is? The solution is n =?

harkovskaia [24]

Answer:

$\frac{1}{3}$ n + 4 = 6

Step-by-step explanation:

Given problem:

4 more than one third of a number n is 6

Unknown:

The equation = ?

Solution of the number = ?

Solution:

let the number = n

So, 4 more than one third of a number;

One third of n = $\frac{1}{3}$ n

4 more;

$\frac{1}{3}$ n + 4

is 6;

$\frac{1}{3}$ n + 4 = 6 (equation of the expression)

Let us now solve the equation:

$\frac{1}{3}$ n + 4 = 6

$\frac{1}{3}$ n = 6 - 4

$\frac{1}{3}$ n = 2

n = 6

5 0

3 years ago

Henry and reiko both used 1 yard of ribbon to make bows. write two different fractions to show that Henry and reiko use the same

Natalka [10]

To show that they used the same amount of ribbon you could take the 1 yard and divided into different sized equal groups.

1 yard divided by four equals 1/4 yard for each of four ribbons.

1 yard divided by six equals 1/6 yard for each of six ribbons.

You could have ribbons the size of 1/4 our yard and 1/6 yard, but you would have different amounts of each.

4 0

3 years ago

Find the rate of change in table plsss helppp!!!!

KonstantinChe [14]

Answer:

it changes by 4

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0

2 years ago