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elixir [45]
3 years ago
8

5x+10=115 what is the value of x​

Mathematics
1 answer:
ololo11 [35]3 years ago
6 0
X = 21

Hope this helped :)
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Janice works for a salary of $2,396 per month. She has federal income tax withheld at the rate of 15 percent, Social Security ta
MAVERICK [17]

Given that Janice monthly salary is $2,396. And she has deductions of federal income tax withheld at the rate 15%, social security tax at the rate of 6.2% and medicare tax at the rate of 1.45% and health insurance premium worth 95$ per month.

Let us calculate total deductions.

Federal income tax = 15% of 2396 = 0.15*2396      =$359.4

Social security tax  = 6.2% of 2396=0.062*2396    =$148.552

Medicare tax           = 1.45% of 2396= 0.0145*2396=$34.742

<u>health insurance premium                                         =$95          </u>

Total deductions                                                        = $637.694

To calculate Janice net pay we have to subtract deductions from monthly salary that is 2396-637.694 = $1758.306

Hence net pay of Janice is $1758.306 per month.

4 0
3 years ago
Tell whether the sequence is arithmetic. If it is, what is the common difference?
denis23 [38]
Idek sorry have a nice day
6 0
3 years ago
Reduce -7/8 + (-1/2) to the simplest form.
ololo11 [35]

Answer:

-1.375

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Hiiya^-^
otez555 [7]

Answer:

I needed thisssss ! Thanks for the points and you too<3

Step-by-step explanation:

3 0
3 years ago
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Which two values of x are roots of the polynomial below?<br> 3x2-3x+1
QveST [7]

Answer:

The roots are

x=\frac{1}{6} [3+ i\sqrt{3}]

x=\frac{1}{6} [3- i\sqrt{3}]

Step-by-step explanation:

we have

3x^2-3x+1

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}

in this problem we have

3x^2-3x+1=0  

so

a=3\\b=-3\\c=1

substitute in the formula

x=\frac{-(-3)\pm\sqrt{-3^{2}-4(3)(1)}} {2(3)}

x=\frac{3\pm\sqrt{-3}} {6}

Remember that

i=\sqrt{-1}

so

x=\frac{1}{6} [3\pm i\sqrt{3}]

The roots are

x=\frac{1}{6} [3+ i\sqrt{3}]

x=\frac{1}{6} [3- i\sqrt{3}]

6 0
3 years ago
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