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3 years ago

11

30 pts. Harrison has 2500 kg of sand delivered to a construction site every day. About how many tons of sand does he have delive

red to the site after 4 weeks? 1 ton≈907.2 kg (Make sure you show work/explain how you arrived at your solution.)

1 answer:

Ymorist [56]3 years ago

6 0

There are 7 days to a week, so 4 weeks = 4 * 7 = 28 days.

28 days x 2,500 kg per day = 70,000 total kgs.

Divide total kgs by kgs per ton:

70,000 total kgs / 907.2 kg per ton = 77.16 tons , round the answer to 77 tons.

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A triangle is formed from the points L(-3, 6), N(3, 2) and P(1, -8). Find the equation of the following lines:

Dima020 [189]

Answer:

Part A) $y=\frac{3}{4}x-\frac{1}{4}$

Part B) $y=\frac{2}{7}x-\frac{5}{7}$

Part C) $y=\frac{2}{7}x+\frac{8}{7}$

see the attached figure to better understand the problem

Step-by-step explanation:

we have

points L(-3, 6), N(3, 2) and P(1, -8)

Part A) Find the equation of the median from N

we Know that

The median passes through point N to midpoint segment LP

step 1

Find the midpoint segment LP

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

L(-3, 6) and P(1, -8)

substitute the values

$M(\frac{-3+1}{2},\frac{6-8}{2})$

$M(-1,-1)$

step 2

Find the slope of the segment NM

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

N(3, 2) and M(-1,-1)

substitute the values

$m=\frac{-1-2}{-1-3}$

$m=\frac{-3}{-4}$

$m=\frac{3}{4}$

step 3

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{3}{4}$

$point\ N(3, 2)$

substitute

$y-2=\frac{3}{4}(x-3)$

step 4

Convert to slope intercept form

Isolate the variable y

$y-2=\frac{3}{4}x-\frac{9}{4}$

$y=\frac{3}{4}x-\frac{9}{4}+2$

$y=\frac{3}{4}x-\frac{1}{4}$

Part B) Find the equation of the right bisector of LP

we Know that

The right bisector is perpendicular to LP and passes through midpoint segment LP

step 1

Find the midpoint segment LP

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

we have

L(-3, 6) and P(1, -8)

substitute the values

$M(\frac{-3+1}{2},\frac{6-8}{2})$

$M(-1,-1)$

step 2

Find the slope of the segment LP

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

L(-3, 6) and P(1, -8)

substitute the values

$m=\frac{-8-6}{1+3}$

$m=\frac{-14}{4}$

$m=-\frac{14}{4}$

$m=-\frac{7}{2}$

step 3

Find the slope of the perpendicular line to segment LP

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-\frac{7}{2}$

so

$m_2=\frac{2}{7}$

step 4

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{7}$

$point\ M(-1,-1)$ ----> midpoint LP

substitute

$y+1=\frac{2}{7}(x+1)$

step 5

Convert to slope intercept form

Isolate the variable y

$y+1=\frac{2}{7}x+\frac{2}{7}$

$y=\frac{2}{7}x+\frac{2}{7}-1$

$y=\frac{2}{7}x-\frac{5}{7}$

Part C) Find the equation of the altitude from N

we Know that

The altitude is perpendicular to LP and passes through point N

step 1

Find the slope of the segment LP

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have

L(-3, 6) and P(1, -8)

substitute the values

$m=\frac{-8-6}{1+3}$

$m=\frac{-14}{4}$

$m=-\frac{14}{4}$

$m=-\frac{7}{2}$

step 2

Find the slope of the perpendicular line to segment LP

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

$m_1*m_2=-1$

we have

$m_1=-\frac{7}{2}$

so

$m_2=\frac{2}{7}$

step 3

Find the equation of the line in point slope form

$y-y1=m(x-x1)$

we have

$m=\frac{2}{7}$

$point\ N(3,2)$

substitute

$y-2=\frac{2}{7}(x-3)$

step 4

Convert to slope intercept form

Isolate the variable y

$y-2=\frac{2}{7}x-\frac{6}{7}$

$y=\frac{2}{7}x-\frac{6}{7}+2$

$y=\frac{2}{7}x+\frac{8}{7}$

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