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Damm [24]
3 years ago
13

A cable company must provide service for 6 houses in a particular neighborhood. They would like to wire the neighborhood in a wa

y to minimize the wiring costs (or distance). What is the minimal length of the network required to span the entire neighborhood? House Distances (yards) 1 to 2 250 1 to 3 400 1 to 4 300 2 to 3 400 2 to 4 400 2 to 5 400 3 to 5 350 3 to 6 450 4 to 5 300 4 to 6 350
Mathematics
1 answer:
8_murik_8 [283]3 years ago
4 0

Answer:

1650 yards

Step-by-step explanation:

Here, we have to find the minimal spanning tree required to span the neighborhood.

We start from house 1. The minimum distance from house 1 to house 2 is 250 yards. Now from 2, we can go to house 3,4 or 5 all having the equal distances  of 400 yard from house 2. So we go to from house 2 to house 3. Now from 3, we go to house 5 which is at a minimum  distance of 350 yards. Now from house 5 we go to house 4 with 300 yards and then from house 4 we go to house 6 which is at 350 yards from 4.

Thus the network is complete and the total distance covered is

= 250 + 400 + 350 + 300 + 350

= 1650 yards

This is the minimum distance by which the neighborhood can be wired.

And the tree is

$1\rightarrow2\rightarrow3\rightarrow5\rightarrow4\rightarrow6$    

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Given log4(3) =0.7925 and log4(5) =1.1610, use log properties to find log4(300).
Ede4ka [16]

Answer:

4.1145

Step-by-step explanation:

You can use the following formula to solve this kind of problems,

log_{a}(b)  +  log_{a}(c)  =  log_{a}(bc)

So, let's start of with log4_(300). Since the question have already gave log4_(3) =0.7925 and log4_(5) =1.1610. It means that you have to split the 300 into the simplest form where almost all of the numbers are 3 and 5.

log_{4}(300)  =  log_{4}(4 \times 5 \times 3 \times 5)  \\  =  log_{4}(4)  + log_{4}(5)  + log_{4}(3)  + log_{4}(5)  \\   = 1 + 1.1610 + 0.7925 + 1.1610 \\  = 4.1145

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10 meters

Step-by-step explanation:

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2 years ago
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Where p is the distance the focus is above the vertex, the equation of a parabola with vertex (h, k) can be written as

... y = 1/(4p)·(x -h)² +k


The vertex is halfway between the focus and directrix. The focus of your parabola is on the y-axis at y=6, and the directrix of your parabola is at y=-6, so the vertex of your parabola is on the y-axis at y=0. That is, the vertex is

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The distance p from the focus at y=6 to the vertex at y=0 is 6 units, so

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