The graph below shows the distance in miles, m, hiked from a camp in h hours. Which hourly interval had the greatest rate of change?

1) hour 0 to hour 1
3) hour 2 to hour 3
2) hour 1 to hour 2
4) hour 3 to hour 4

The greatest rate of change between the interval 0 to 1 hour.

Given that,

The graph below shows the distance in miles, m, hiked from a camp in h hours.

We have to determine,

Which hourly interval had the greatest rate of change?

According to the question,

The greatest rate of change between the interval is determined by the rate of change between two points refers to the vertical change divided by the horizontal change.

1. The rate of change between 0 to 1 hours is,

The rate of change between 0 to 1 hour is 2.

2. The rate of change between 2 to 3 hours is,

The rate of change between 2 to 3 hours is 1.

3. The rate of change between 1 to 2 hours is,

The rate of change between 1 to 2 hours is 1.5.

4. The rate of change between 3 to 4 hours is,

The rate of change between 2 to 3 hours is 1.

Hence, The required greatest rate of change between the interval 0 to 1 hour.

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Option 1) is correct

Step-by-step explanation:

Given: The graph shows the distance in miles, m, hiked from a camp in h hours.

To find: hourly interval that had the greatest rate of change

Solution:

Rate of change between two points refers to the vertical change divided by the horizontal change.

For hour 0 to hour 1:

rate of change =

For hour 2 to hour 3:

rate of change =

For hour 1 to hour 2:

rate of change =

For hour 3 to hour 4:

rate of change =

So, rate of change is greatest in interval: hour 0 to hour 1

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.