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Lana71 [14]
3 years ago
11

A price of a phone is marked down from 350 to 287 for a sale. the following week the phone is marked down again but the same per

cent as during the week before. how much lower than the original price is the price after the second markdown?
A) 51.66
B) 114.66
C) 126
D) 235.34
Mathematics
1 answer:
ad-work [718]3 years ago
7 0

Answer: D. 235.34

To answer this question we must first find the percentage of decrease.

To do this we must subtract 287 from 350

to find the difference. -63

Next, we must divide -63 by the original number, 350, to find -0.18.

Now we will multiply -0.18 by 100 to find the percentage.

So, the decrease is by -18%

Because -18% of 287 is 51.66 this will be the amount of money we subtract.

So, 287 - 51.66

235.34 is the price  after the second markdown.

I hope you found this helpful!

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\large\boxed{\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\dfrac{\sqrt5}{x^2y}}

Step-by-step explanation:

\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{55}{11}\cdot\dfrac{x^7}{x^{11}}\cdot\dfrac{y^6}{y^8}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\=\sqrt{5x^{7-11}y^{6-8}}=\sqrt{5x^{-4}y^{-2}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\=\sqrt5\cdot\sqrt{x^{-4}}\cdot\sqrt{y^{-2}}=\sqrt5\cdot\sqrt{x^{(-2)(2)}}\cdot\sqrt{y^{(-1)(2)}}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt5\cdot\sqrt{(x^{-2})^2}\cdot\sqrt{(y^{-1})^2}\qquad\text{use}\ \sqrt{a^2}=a

=\sqrt5\cdot x^{-2}\cdot y^{-1}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\to a^{-1}=\dfrac{1}{a}\\\\=\sqrt5\cdot\dfrac{1}{x^2}\cdot\dfrac{1}{y}=\dfrac{\sqrt5}{x^2y}

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