Alana and kelsa are training for a marathon. to prepare for this marathon they have been training and tracking their progress periodically.
in the first week of training alana ran an average of 12 minutes per mile. later, in week five of training she ran an average of 6 minutes per mile.
in the first week of training kelsa ran an average of 10 minutes per mile. later, in week eleven of training she ran an average of 5 minutes per mile.
assuming that alana and kelsa continue to train and improve their times at the same rate your task is to determine which week they will have the same average minutes per mile. we will assume that the relationship is linear as they will be training for a maximum of 25 weeks. to complete this task follow the steps below.
1. determine the equation of a line in standard form that represents alana’s training progress. her progress corresponds to the points (1, 12) and (5,6).

2. determine the equation of a line in standard form that represents kelsa’s training progress. her progress corresponds to the points (1, 10) and (11, 5)

3. solve the system of equations (you must show all your work to receive full credit).

4. in which week will alana and kelsa have the same average minutes per mile?

5. if alana and kelsa continue to train until week 25, what will their times be?

6. do you believe a linear model best represents the relationship of the time of the runners and the weeks that passed? (hint: look at question 5). what do you think this says about problems in the real world? justify your thoughts in 3-4 sentences.

Alana's training experience in (weeks, minutes per mile) is (1, 12) and (5, 6).

Kelsa's training experience is (1, 10), (11, 5).

Please note that in parts 1 and 2 a "standard form" linear equation has a positive leading coefficient and the numbers have no common factor.

1. A linear equation can be written from two points (x1, y1) and (x2, y2) as ...

... (y2 -y1)(x -x1) -(x2 -x1)(y -y1) = 0

Using the given points in order, this becomes

... (6-12)(x -1) -(5 -1)(y -12) = 0 . . . . above equation with values filled in

... -6x +6 -4y +48 = 0 . . . . . . . . . . eliminate parentheses

... 3x +2y -27 = 0 . . . . . . . . . . . . . divide by -2 to make x-coefficient positive

... 3x +2y = 27 . . . . . . . . . . . . . . . . Alana's training experience

2. In similar fashion, we can write the equation for Kelsa's experience.

... (5 -10)(x -1) -(11 -1)(y -10) = 0

... -5x +5 -10y +100 = 0 . . . . . . eliminate parentheses

... x +2y -21 = 0 . . . . . . . . . . . . . divide by -5

... x +2y = 21 . . . . . . . . . . . . . . . . Kelsa's training experience

3. We note that both equations have the term 2y, so we can subtract the second equation from the first to eliminate that term:

... (3x +2y) -(x +2y) = (27) -(21)

... 2x = 6 . . . . . . . simplify

... x = 3 . . . . . . . . divide by 2

... 3·3 +2y = 27 . . . substitute x=3 into Alana's equation

... 2y = 18 . . . . . . . . subtract 9

... y = 9 . . . . . . . . . . divide by 2

The solution is (weeks, minutes) = (x, y) = (3, 9).

4. The solution of part 3 tells us both have an average time of 9 minutes per mile in week 3 of training.

5. Substituting x=25 into the equations of parts 1 and 2, we get ...

... 3·25 +2y = 27

... y = (27 -75)/2 = -24 . . . . Alana's time per mile at week 25

... 25 + 2y = 21

... y = (21 -25)/2 = -2 . . . . . Kelsa's time per mile at week 25

6. A linear model makes no sense in this context. At some point a linear equation for minutes per mile will always have a value of 0. We know that marathon runners will never have a time of zero minutes per mile. Under the best of conditions, we expect training to improve time toward some asymptotic limit determined by body mechanics. A linear model is completely inappropriate for describing such training effects. Not every real-world problem is appropriately represented by a linear equation.

b

step-by-step explanation:

calculate the scale factor as the ratio of corresponding sides, image to original, that is

scale factor = = = = 1

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3