1) The Boateng family has a sickening house whose layout is shown above. They would like to renovate just bedrooms #2 and 3. Because you
are a contractor, you receive 20% off from Home Depot whenever you
make any purchases. If Home Depot sells your client's ideal wool carpet
for $8.99 per square foot, how much will it cost to carpet both
bedrooms? (Keep in mind that 10'8" would be written as a decimal
approximation of 10.67 ft. Think about why this is.)
a) How many total square feet of carpeting will you need?

The cost will be $1662.136  

The total square ft of carpeting will be 231.1122ft^2    

Step-by-step explanation:

For the bedrooms,  the dimension from the image is shown as(see attahced)

Bedroom #2= 10"8' x10"4'

Bedroom #3= 10"8' x11"4'

let us convert all dimension to feet and solve for the area of the bedrooms

Bedroom #2= 10"8' x10"4'

Bedroom #2=10.67x10.33feet

Area of Bedroom #2= 110.2211ft^2

Bedroom #3 = 10"8' x11"4'

Bedroom #3 =10.67x11.33feet

Area of Bedroom #3= 120.8911ft^2

The total area of both rooms is

Total area=110.2211+120.8911

Total area=231.1122ft^2

now the cost of one square floor s $8.99

hence the cost of 231.1122ft^2 will be

Cost of carpet = 231.1122*8.99

Cost of carpet = $2,077.67

Given that, because you are a contractor, you will receive 20% off

therefore it will cost

= (20/100)*2,077.67

=0.2*2,077.67

=$415.534

The cost will now be  $2,077.67-$415.534

= $1662.136

  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.

bottom left

step-by-step explanation:

the x's and the y's both go up the same amount each time, therefore it will be a linear function because it will go in a straight line when drawn on a graph.

hope this !