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2 years ago

Single variable linear equation that has one solution

Mathematics

1 answer:

Tcecarenko [31]2 years ago

3 0

Is the variable segment negative or positive determining for the answer

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The slope of a hill next to a highway decreases 2 vertical feet for every 3 horizontal feet away from the highway.

Novay_Z [31]

Slope is the change in Y ( vertical) over the change in X( horizontal).

The slope would be -2 /3

You would multiply the horizontal distance by the slope:

-2/3 x 12 = -8

The answer would be D. v(h)=-2/3h, and the vertical distance is -8 feet.

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levacccp [35]

Answer:

2.2

Step-by-step explanation:

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Yakvenalex [24]

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A city planner designs a park that is a quadrilateral with vertices at J(-3,1), K(1,3), L(5,-1), and M(-1,-3). There is an entra

Ksivusya [100]

The Quadrilateral is JKLM,

let $M_{JK}, M_{KL}, M_{LM}, M_{JM},$ be the midpoints of JK, KL, LM and JM respectively.

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Given any 2 point P(m,n) and Q(k,l),<span>

the coordinates of the midpoint of the line segment PQ are given by the formula:

$M_{PQ}=( \frac{m+k}{2} ,
\frac{n+l}{2})$ , </span>

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thus the coordinates of points $M_{JK}, M_{KL}, M_{LM}, M_{JM},$

are as follows:

$M_{JK}= (\frac{-3+1}{2}, \frac{1+3}{2})=(-1,2), \\\\M_{KL}= (\frac{1+5}{2}, \frac{3-1}{2}=(3,1), \\\\M_{LM}= (\frac{5-1}{2}, \frac{-1-3}{2})=(2, -2),\\\\ M_{JM}= (\frac{-3-1}{2}, \frac{1-3}{2})=(-2,-1)$

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The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

$|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}
}$ </span>

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thus the distances connecting the opposite entrances can be calculated as follows:

$|M_{JK},M_{LM}|= \sqrt{ (-1-2)^{2} + (2-(-2))^{2} }= \sqrt{9+16}=5$

$|M_{KL}M_{JM}|= \sqrt{ (3-(-2))^{2} + (1-(-1))^{2}}= \sqrt{25+4}= \sqrt{29}=5.39$

Thus the total distance of the paths joining the opposite entrances is

5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)

Answer: 104 m

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2 years ago