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Law Incorporation [45]
3 years ago
15

Margie has these times recorded on her time sheet for the week. How many hours did she work?

Mathematics
2 answers:
Mumz [18]3 years ago
7 0
The answer is
A) 47
kobusy [5.1K]3 years ago
4 0

Answer:

Option A. 47 hours

Step-by-step explanation:

The given table is about the times recorded on the time sheet for the week.

We have to calculate the total time she worked for.

From 7:05 to 3:58

If 7:05 is a.m. then 3:58 will be 15:58 pm.

By subtraction of these timings we can get the number of hours she has worked.

15:58 - 07:05 = 8:53 (8 hours 53 minutes)

From 7:30 a.m. to 17:01 pm

17:01 - 07:30 = 9:31 (9 hours 31 minutes)

From 8:01 a.m. to 17:32 pm

17:32 - 08:01 = 09:31 ( 9 hours 31 minutes)

From 07:22 a.m. to 16:46 pm

16:46 - 07:22 = 09:24 (9 hours 24 minutes)

From 6:55 am to 16:31 pm

16:31 - 06:55 = 9:36 (9 hours 36 minutes)

Therefore, total time she worked = 8:53 + 9:31 + 9:31 + 9:24 + 9:36 = 44:175 (44 hours 175 minutes)

Now we will convert minutes to hours

175 minutes = 2 hours 55 minutes

So total working hours = 44 hours + 2 hours 55 minutes = 46 hours 55 minutes ≈ 47 hours

Therefore, Option A. 47 hours is the answer.

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Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6
kenny6666 [7]

Answer:

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

Step-by-step explanation:

Given the simultaneous equations

y=9-x

y\:=\:2x^2\:+\:4x\:+\:6

Subtract the equations

y=9-x

-

\underline{y=2x^2+4x+6}

y-y=9-x-\left(2x^2+4x+6\right)

\mathrm{Refine}

x\left(2x+5\right)=3

\mathrm{Solve\:}\:x\left(2x+5\right)=3

2x^2+5x=3        ∵ \mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x

\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}

2x^2+5x-3=3-3

\mathrm{Solve\:with\:the\:quadratic\:formula}

\mathrm{Quadratic\:Equation\:Formula:}

\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\

x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}

  =\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{2\cdot \:2}

  =\frac{-5+\sqrt{49}}{4}

  =\frac{-5+7}{4}

  =\frac{2}{4}

  =\frac{1}{2}

Similarly,

x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3

\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}

x=\frac{1}{2},\:x=-3

\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}

\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12

\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}

\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}

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Answer:

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a^{2}

Step-by-step explanation:

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