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Helga [31]
3 years ago
12

1. Sarah owns a small business. There was a loss of S19 on Thursday and a loss of $12 on Friday. On Saturday there was a loss of

$11, and on Sunday there was a
profit of $15. Find the total profit or loss for the four days.
$19 profit
$35 loss
$57 profit
S27 loss
Mathematics
1 answer:
Montano1993 [528]3 years ago
8 0

Sarah had a loss of $27 for the four days.

Step-by-step explanation:

Given,

Loss on Thursday = $19

Loss on Friday = $12

Loss on Saturday = $11

Profit on Sunday = $15

Lets suppose loss as negative amount and profit as positive amount.

Adding all the loss and profit = (-19)+(-12)+(-11)+(15)

Total = -19-12-11+15

Total = -27\\

As we supposed negative amount as loss, therefore,

Sarah had a loss of $27 for the four days.

Keywords: Profit, loss

Learn more about profits at:

  • brainly.com/question/9998060
  • brainly.com/question/9967230

#LearnwithBrainly

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Answer:

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Step-by-step explanation:

The square root of 3 is irrational, because 3 is not a perfect square.

For example, 4 is a perfect square, so the square root (2) will be rational.

The square root of 3 is 1.73205080757,

and we can round this to 1.7 for the tenths place

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What is the median for 252,210,264,278,208,295,248,257,284,271....expressed to the nearest tenth.
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\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)



\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\

4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
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x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}
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