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zepelin [54]
3 years ago
15

Ou are setting up a zip line in your yard. You map out your yard in a coordinate plane. An equation of the line representing the

zip line is y=-8/3+8
. There is a tree in your yard at the point (3,16) . Each unit in the coordinate plane represents 1 foot. Approximately how far away is the tree from the zip line? Round your answer to the nearest tenth.
Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0

9514 1404 393

Answer:

  5.6 feet

Step-by-step explanation:

Assuming your zip line equation is supposed to be ...

  y = -8/3x +8

we can rewrite it to general form:

  8/3x +y -8 = 0

  8x +3y -24 = 0 . . . . multiply by 3

Now, we can use the formula for the distance from a point to a line:

  d = |ax +by +c|/√(a²+b²) . . . . . distance from (x, y) to ax+by+c=0

Filling in the values we know, we have ...

  d = |8(3) +3(16) -24|/√(8² +3²) = 48/√73 ≈ 5.6

The distance from the tree to the zip line is about 5.6 feet.

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Answer:

^{(x-y)}C_{10}=\frac{(x-y)!}{10! \times (x-y-10)!}

Step-by-step explanation:

Total flavors Sammy initially had = x

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After throwing away y flavors, the number of flavors Sammy will be left with = x - y

He needs to make 10-flavor bags from these (x - y) flavors. In order words he needs to chose 10 flavors for each bag from(x - y) flavors. The order of selection is not important here, so this is a problem of combinations. Also since we have to make selections or small groups, this also indicates that we have to use combinations.

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Using the values in this formula, we get:

^{(x-y)}C_{10}=\frac{(x-y)!}{10! \times (x-y-10)!}

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