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2 years ago

6

Diane has been keeping an eye on an emerald ring she likes. Its original price was $22,090. After being marked down for a cleara

nce sale, it is now $13,254. By what percent has the price of the ring gone down?

1 answer:

xz_007 [3.2K]2 years ago

8 0

Answer:

40%

Step-by-step explanation:

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PLS I NEED HELP THIS IS MY LAST QUESTION PLS

Keith_Richards [23]

Answer:

82cm have a great day

Step-by-step explanation: have a great day :)

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2 years ago

Write thThe bookstore is selling a series of 4 books for $97.50. What is the unit price for one book? Round your answer to the n

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3 years ago

(3x)^2 write in expanded form use parentheses to indicate multiplication

Korolek [52]

Expand (3x)^2

When you see ^2, it means that the term(s) inside the parenthesis should be copied once, and multiplied.

(3x)^2 = (3x)(3x)

(3x)(3x) is your expanded form

<em>~Rise Above the Ordinary</em>

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3 years ago

The sum of three consecutive even integers is -114.

Alika [10]

The sum of three consecutives is b

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3 years ago

How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

7 0

3 years ago