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.
Hi
According to Pythagoras :
The square of hypotenuse is the sum if the square if the two other side
so : 7^2 + X^2 = 25^2
49 + X^2 = 625
X^2 = 625-49
X^2 = 576
this equation as two roots . As a triangle cannot have a negative side value, I will only take positive value
so X = V576 =24
Answer:
8/12
Step-by-step explanation:
2/3 x 4= 8/12
Answer: 24
Step-by-step explanation:
3/4 divides by 1/7 =
copy dot flip
3/4 * 7/1
21/4
4 goes into 21 5 times with 1 left over
5 1/4