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aleksandr82 [10.1K]
3 years ago
11

During a bite challenge rattlers how to collect various colored ribbons each 1/2 mile they collect a red ribbon each 1/8 mile th

ey collect a green ribbon and each 1/4 mile they collect a blue ribbon which colors or ribbons will be collected
Mathematics
2 answers:
tino4ka555 [31]3 years ago
7 0
1/2×4=4/8
1/8=1/8
1/4×2=2/8
so 4/8+2/8=6/8
and 6/8+1/8=7/8

so 7/8 ribbons will be collected
AleksAgata [21]3 years ago
4 0

Question

During a bike challenge, riders have to collect various colored ribbons. Each 1/2 mile they collect a red ribbon, each 1/8 mile they collect a green ribbon, and each 1/4 mile they collect a blue ribbon. Which colors of ribbons will be collected at the 3/4 mile marker?

Answer:

A green and a blue ribbon will be collected at the  \frac{3}{4} mile marker.

Step-by-step explanation:

<em>Topic: Factors and Multiples</em>

<em></em>

To solve this, we'll write out the first few multiples of the distances where the collection of ribbons occurs:

Given that each \frac{1}{2} mile they collect a red ribbon; The multiples for the Red Ribbon are as follows (by adding \frac{1}{2} at each progression):

Red: \frac{1}{2}  -> {1}{}  -> \frac{3}{2}  -> {2}{}  -> \frac{5}{2}

Given that each \frac{1}{8} mile they collect a green ribbon; The multiples for the Green Ribbon are as follows (by adding \frac{1}{8} at each progression):

Green: \frac{1}{8}  -> \frac{1}{4}  -> \frac{3}{8}  -> \frac{1}{2}  -> \frac{5}{8} -> \frac{3}{4}  -> \frac{7}{8}  -> {1}

Given that each \frac{1}{4} mile they collect a blue ribbon; The multiples for the Blue Ribbon are as follows (by adding \frac{1}{4} at each progression):

Blue: \frac{1}{4} -> \frac{1}{2}  -> \frac{3}{4}  -> {1}

Since , \frac{3}{4} is a multiple of both  \frac{1}{8} and \frac{1}{4} , then a green and a blue ribbon will be collected at the  \frac{3}{4} mile marker.

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Answer:

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Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

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∇f = (2x, 2y, 2z )

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Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

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                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

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however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

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x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

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Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

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