Given that the projectiles have been given by the functions:
H(t)=-16t^2+96t+80 and G(t)=31+32.2t
Part A:
The tables for the functions will be as follows:
t 2 3 4 5
H(t) 208 224 208 160
t 2 3 4 5
G(t) 95.4 127.6 159.8 198
The solution is found between points:
4th second and 5th second
i] It's between this point that the graph H(t) is has reached the maximum point and it's now turning. So the points of H(t) are nearing points for G(t).
Part B]
The solution in part A implicates the times at which the projectiles were at the same height and the time at which they were at the same heights.
Answer:
3.2 height
Step-by-step explanation:
54-31.6=22.4
15-8=7
22.4/7=3.2
Answer:
Step-by-step explanation:
Given
So:
--- May
--- June
Required
Express as a function
Start by calculating the slope (m)
Simplify
The equation is:
Take LCM
Answer:
A.) gf(x) = 3x^2 + 12x + 9
B.) g'(x) = 2
Step-by-step explanation:
A.) The two given functions are:
f(x) = (x + 2)^2 and g(x) = 3(x - 1)
Open the bracket of the two functions
f(x) = (x + 2)^2
f(x) = x^2 + 2x + 2x + 4
f(x) = x^2 + 4x + 4
and
g(x) = 3(x - 1)
g(x) = 3x - 3
To find gf(x), substitute f(x) for x in g(x)
gf(x) = 3( x^2 + 4x + 4 ) - 3
gf(x) = 3x^2 + 12x + 12 - 3
gf(x) = 3x^2 + 12x + 9
Where
a = 3, b = 12, c = 9
B.) To find g '(12), you must first find the inverse function of g(x) that is g'(x)
To find g'(x), let g(x) be equal to y. Then, interchange y and x for each other and make y the subject of formula
Y = 3x + 3
X = 3y + 3
Make y the subject of formula
3y = x - 3
Y = x/3 - 3/3
Y = x/3 - 1
Therefore, g'(x) = x/3 - 1
For g'(12), substitute 12 for x in g' (x)
g'(x) = 12/4 - 1
g'(x) = 3 - 1
g'(x) = 2.
