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Bad White [126]
2 years ago
14

A walk-a-thon was held at a local middle school. These are the average distances and times for three walkers at different times

during the walk-a-thon. Lakeisha: distance- 7/8 and time- 18 minutes Baydan: distance-9/10 and time- 22 minutes Madison: distance-4/5 and time- 17 minutes. If the total distance of the route was 2 miles who completed the route first? How long did it take them if they stayed at a constant rate?
Mathematics
1 answer:
Mekhanik [1.2K]2 years ago
6 0

The person that completed the 2 miles route first is Lakeisha and he did it in a time of 41.15 minutes

<h3>How to find total time from distance?</h3>

The average distances and times are as follows;

Lakeisha: distance = 7/8; Time = 18 minutes

Baydan: distance = 9/10; Time = 22 minutes

Madison: distance = 4/5; Time = 17 minutes

Formula for speed is;

Speed = Distance/Time

Speed of Lakeisha = (7/8)/18 = 0.0486 Miles per minutes

Speed of Bayden = (9/10)/22 = 0.041 Miles per minutes

Speed of Madison = (4/5)/17 = 0.047 Miles per minutes

Thus, the person that will complete the 2 miles route first is the person with the fastest speed which is Lakeisha.

Time he will take to complete the 2 miles route = 2/0.0486 = 41.15 minutes

Read more about time from distance at; brainly.com/question/17146782

#SPJ1

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The simplified answer is \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}.

<u>Step-by-step explanation:</u>

$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term (\frac{3x+y}{3x+y}) and 2nd term by (\frac{z+2x}{z+2x})

= \frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= \frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}

Now, the denominators are same, you can subtract it.

= \frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

= \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

Thus the simplified solution is  \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

4 0
4 years ago
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Answer: Choice C) 115 degrees

--------------------------------

Explanation:

The three angles are 42, 73 and 65 degrees. Let's call them x,y, and z respectively
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To find an exterior angle for x, we subtract the given angle x from 180
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4 years ago
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