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anygoal [31]
3 years ago
11

Three runners competed in a race.

Mathematics
1 answer:
hram777 [196]3 years ago
6 0

Answer:

runner B

Step-by-step explanation:

its the only  one on the line

runner C is tall like a mountan

and runner A is not anywhere close to a line

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Nicole has used 6 feet of ribbon.This represents 3/8 of the total amount of the ribbon she started with. How much ribbon did Nic
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3/8x=6. x=6*8/3. x=2*8. x=16. Nicole started with 16 feet of string.
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Solve #14 for x and y plsss I WILL GIVE BRAINLY
frez [133]

7x - 44 = 4x + 4

Add both sides 44

7x - 44 + 44 = 4x + 4 + 44

7x = 4x + 48

Subtract both sides 4x

7x - 4x = 4x - 4x + 48

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<h2>x = 16 </h2>

______________________________

Suppose that the angle between 39 and 4x + 4 is t degrees as you know they make a straight line together thus :

39 + t + 4x + 4 = 180

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_____________________________

8y - 43 = 4x + 4 + t

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8y - 43 = 68 + 73

8y - 43 = 141

Add both sides 43

8y - 43 + 43 = 141 + 43

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Divide both sides by 8

8y ÷ 8 = 184 ÷ 8

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3 years ago
Julie earned 66 on the test. Describe how the scores of Julie’s classmates compared to Julie’s score.
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<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
3 years ago
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