A segment is m units long. Find the distance between the midpoints of the first and the last parts in each case. a The segment is divided into three equal parts

Line can be divided into segments.

The distance between the two midpoints is 2m/3

The length of the segment is given as:

The length of each segment, after dividing them into 3 segments is:

The midpoint of the first segment is:

The midpoint of the third segment is the sum of the first two segments, and the half a segment

So, we have:

So, the distance between the two midpoints is:

Hence, the distance between the two midpoints is 2m/3

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(2/3)m units

Step-by-step explanation:

The segment is m units long. After dividing it into 3 three equal parts, each part is m/3 units long. The midpoint of the first part is located at (m/3)/2 = m/6 units from the beginning of the segment. The midpoint of the last part is located at (2/3)m + m/6 = (5/6)m units from the beginning of the segment. Then, the distance between the midpoints of the first and the last parts is (5/6)m - m/6 = (2/3)m units

play black ops 4 answer a

step-by-step explanation:

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)