At the time the rocket hits the ground h=0, given that h=-16t²+320t+32
when h=0, our equation will be:
-16t²+320t+32=0
solving the above by completing square method we proceed as follows;
-16t²+320t+32=0
divide though by -16 we get
t²-20t-2=0
t²-20t=2
but
c=(-b/2a)^2
c=(20/2)^2
c=100
hence:
t²-20t+100=100+2
(t-10)(t-10)=102
√(t-10)²=√102
t-10=√102
hence
t=10+/-√102
t~20.1 or -0.1
since it must have taken long, then the answer is 20.1 sec
Answer:
procedure always produces 6
Step-by-step explanation:
Let 'n' be the unknown number
Add 4 to the number : n+4
multiply the sum by 3.
multiply the sum n+4 by 3

Now subtract 6, so we subtract 6 from 3n+12

finally decrease the difference by the tripe of the original number
triple of original number is 3n

so the procedure always produces 6
I would guess the answer to be Tension, if not for the extra 'X'....
or
Extension, if you add another 'E'...
Answer:
119.45 m
Step-by-step explanation:
Given:
When angle of elevation of the sun changes from 58° to 36° the length of shadow of a pole increases by 90 m.
To find:
Length of pole = ?
Solution:
Kindly refer to the attached image.
represents the 1st angle of elevation of sun i.e. 58°
represents the 2nd angle of elevation of sun i.e. 36°
Change in shadow is represented by CD = 90 m
Let height of pole, AB =
m
Let side BC =
m
Now, let us apply tangent rules in
one by one:


Putting value of
using equation (1):

119.45 m is the height of pole.