7. RDP stands for Reconstruction and Development Programme. A timetable is given for the construction of a RDP house with one bedroom, a lounge, Kitchenette and toilet
Working day
number
Project
Sto 6
Site preparation
Foundation construction
Building walls
Installation of windows and
8 to 12
10 to 12
ons
Roofing
12 to lo
Electrical
12 to 10
Plumbing
17 to 20
Painting
Flooring
Hook up to water main
Hook up to sewer or installation of
septic system
a) Within how many days will the house be completed?
b) In how many days must the walls be built?
. What has to be done at the same time as the roofing?
d) How many days are available to paint the house?
The site preparation of a house starts on Thursday 5 April
. Bullders, plumbers and electricians work​

a net is a diagram that shows a “flattened” view of a solid. in a net, each face and base is shown with all of its dimensions. a net can also serve as a pattern to build a three-dimensional solid.

a pyramid is a three-dimensional object with a base that is a polygon and triangular faces that meet at one vertex.

" the net of a triangular pyramid consists of four triangles. the base here we are given of the pyramid is a equilateral   triangle, and the lateral faces are also triangles "

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

if you need just tell me and i will come to the rescue.

step-by-step explanation:

its an acute right triangle

step-by-step explanation: