Find the midpoint of pq
p(4 1/3, 3 1/6), q(-2 1/5, 3 2/3)

Step-by-step explanation:

To find the midpoint, you must average the x's then average the y's.

So the x's are and .

The y's are and .

To find an average of a data with 2 elements you must add the two elements then take that sum and divide it by 2.

So this is what we will do with the x's then the y's.

So first x:

Write as improper fractions:

Division by 2 is the same as multiplying by 1/2:

The least common multiple of 6 and 10 is 30.  We will multiply first fraction by 5/5 and 3/3 for the second so we can have the same denominator.

Now that the bottoms are the same we can write as one fraction:

Simplify:

If you want the answer as a mix fraction like your question began as we can do that by first figuring out how many 30's are in 32 and what is the remainder of that division.

There is one 30 and 32 and so 32-30=2 is the remainder.

Reduce by dividing top and bottom by 2:

Now for y:

Write the mix fractions as improper fractions:

Dividing by 2 is the same as multiplying by 1/2:

The least common multiple of 12 and 6 is 12 so I'm going to multiply the second fraction by 2/2 so the denominators will be the same:

Since the denominators are the same we can write as a single fraction:

How many 12's are in 41? 3 and so the remainder is 41-3(12)=41-36=5.

So the midpoint is:

.

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.

my guess would be b and d, but i didn’t actually get out the pen and paper. lol