Fredric leads a team of hikers for a full-day hike. the total elevation gain during the hike is 2,100 feet. all of the hikers have to pass two checkpoints before they reach the peak. the elevation gain from the starting point to checkpoint 1 is 100 feet less than double the elevation gain from checkpoint 2 to the peak. the elevation gain from checkpoint 1 to checkpoint 2 is the mean of the elevation gain from the start to checkpoint 1 and the elevation gain from checkpoint 2 to the peak. let x represent the elevation gain from the starting point to checkpoint 1, y represent the elevation gain from checkpoint 1 to checkpoint 2, and z represent the elevation gain from checkpoint 2 to the peak. which augmented matrices represent the context of this scenario?

1    1    1    2100

1    0   -2  -100

-1   2   -1    0

Step-by-step explanation:

x: elevation gain from the starting point to checkpoint 1 (in feet)

y: elevation gain from checkpoint 1 to checkpoint 2 (in feet)

z: elevation gain from checkpoint 2 to the peak (in feet)

The total elevation gain during the hike is 2,100 feet:

x + y + x = 2100 (eq. 1)

The elevation gain from the starting point to checkpoint 1 is 100 feet less than double the elevation gain from checkpoint 2 to the peak:

x = 2z - 100

x + 0y - 2z = - 100 (eq. 2)

The elevation gain from checkpoint 1 to checkpoint 2 is the mean of the elevation gain from the start to checkpoint 1 and the elevation gain from checkpoint 2 to the peak:

y = (x + z)/2

2y = x + z

-x +2y -z = 0 (eq. 3)

From the system of equations 1, 2 and 3 we can get the following augmented matrix:

1    1    1    2100

1    0   -2  -100

-1   2   -1    0

where the first column are the coefficients of x, the second column the coefficients of y, the third column the coefficients of z, and the last column the constants.

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.

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%