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3 years ago

1/8 (16k-24)+1/5 (2+10k)

1 answer:

3 0

¹/₈(16k - 24) + ¹/₅(2 + 10k)
¹/₈(16k) + ¹/₈(-24) + ¹/₅(2) + ¹/₅(10k)
2k + (-3) + ²/₅ + 2k
2k - 3 + ²/₅ + 2k
2k + 2k - 3 + ²/₅
4k - 2³/₅

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25 decreased four times n

disa [49]

The answer is 4n-25
4 times n is 4 n

and 25 decreased or subtracted from the 4n
thus the answer is 4n-25

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3 years ago

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A television Mark 1600 was sold for 1200 what percent discount was given on that television

adoni [48]

Answer:

25%

Step-by-step explanation:

Discount = 1600 - 1200 = 400

Percentage discount

$= \frac{400}{1600} \times 100 \\ \\ = \frac{400}{16} \\ \\ = 25\%$

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2 years ago

The cost to attend a Wildcat sports event is: Adult: $7 Student: $5 The total sales was $700.

hram777 [196]

Answer:

7x+5y=700

91 student tickets

Step-by-step explanation:

7x+5y=700

7(35)+5y=700

245+5y=700

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3 years ago

There are 40 markers in a bag. Of the 10 scented markers, 6 are permanent markers. Half of the unscented markers are erasable.

Vera_Pavlovna [14]

Answer:

K12 ANSWER

Step-by-step explanation:

7 0

3 years ago

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Please help radical expression dividing

77julia77 [94]

$\bf \cfrac{\sqrt[4]{63}}{4\sqrt[4]{6}}\qquad 
\begin{cases}
63=3\cdot 3\cdot 7\\
6=2\cdot 3
\end{cases}\implies \cfrac{\sqrt[4]{3\cdot 3\cdot 7}}{4\sqrt[4]{2\cdot 3}}\implies \cfrac{\underline{\sqrt[4]{3}}\cdot \sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}\cdot \underline{\sqrt[4]{3}}}
\\\\\\
\cfrac{\sqrt[4]{3}\cdot \sqrt[4]{7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{3\cdot 7}}{4\sqrt[4]{2}}\implies \cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}$

$\bf \textit{now, rationalizing the denominator}\\\\
\cfrac{\sqrt[4]{21}}{4\sqrt[4]{2}}\cdot \cfrac{\sqrt[4]{2^3}}{\sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21}\cdot \sqrt[4]{8}}{4\sqrt[4]{2}\cdot \sqrt[4]{2^3}}\implies \cfrac{\sqrt[4]{21\cdot 8}}{4\sqrt[4]{2\cdot 2^3}}\implies \cfrac{\sqrt[4]{168}}{4\sqrt[4]{2^4}}
\\\\\\
\cfrac{\sqrt[4]{168}}{4\cdot 2}\implies \cfrac{\sqrt[4]{168}}{8}$

and is all you can simplify from it.

so... all we did, was rationaliize it, namely, "getting rid of the pesky radical at the bottom", we do so by simply multiplying it by something that will raise the radicand, to the same degree as the root, thus the radicand comes out.

6 0

3 years ago