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GalinKa [24]
3 years ago
6

Glowmart marked down it's stock 32%. What will be the sale price of a $60 jacket​

Mathematics
2 answers:
Igoryamba3 years ago
5 0

Answer:

Sale price = $40.80

Step-by-step explanation:

The original price = $60

After mark down of 32% is applied,

The selling price becomes 100% - 32% = 68%

The selling price = 68% of $60 = \frac{68}{100} × $60 = $40.80

tatuchka [14]3 years ago
4 0

Answer: $40.80

Explanation: 60*.68=40.8

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When Javier and his father went to dinner, the bill was $34. If they tipped the waiter 20%, how much did they spend in total?
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Not sure if this is right, but I got $40.8.

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Solve irrational equation pls
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\hbox{Domain:}\\&#10;x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\&#10;x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\&#10;x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\&#10;(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\&#10;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)\\&#10;x\in(-\infty,-2\rangle\cup\langle3,\infty)


&#10;\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\&#10;x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\&#10;2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\&#10;\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\&#10;(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\&#10;(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\&#10;4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\
&#10;4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\&#10;(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\&#10;(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\&#10;(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\&#10;(x-1)^2(3x^2-28)=0\\&#10;x-1=0 \vee 3x^2-28=0\\&#10;x=1 \vee 3x^2=28\\&#10;x=1 \vee x^2=\dfrac{28}{3}\\&#10;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\\&#10;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}}\}\\&#10;\boxed{\boxed{x=1}}
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