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dybincka [34]
2 years ago
9

If the data (x+1), (2x+1), (x+7), and (3x+4) are in ascending order and the median is 16, then find the value of 'X'.​

Mathematics
1 answer:
Sonbull [250]2 years ago
7 0

Answer:

sum \:  = (x + 1) + (2x + 1) + (x + 7) + (3x + 4) \\  = 7x + 13 \\ median = ( \frac{sum}{2}  + 1) \\ 16 = ( \frac{7x + 13}{2}  + 1) \\ 7x + 13 = 14 \\ 7x = 1 \\ x =  \frac{1}{7}

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Eddie the elf made 13 toys one day in the workshop. some were dolls and the rest were action figures. the number of action figur
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3 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0
2 years ago
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jonny [76]

Answer:

SSS

Step-by-step explanation:

ST = YA, SY = TA Given

AS = SA Reflexive property

ΔSTA ≅ ΔAYS SSS

8 0
3 years ago
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