Answer:4.5 x 2
Step-by-step explanation:
4.5 x 2 = 9
Hope it helps
Answer:
Hence after 3.98 sec i.e 4 sec Object will hit the ground .
Step-by-step explanation:
Given:
Height= 6 feet
Angle =28 degrees.
V=133 ft/sec
To Find:
Time in seconds after which it will hit the ground?
Solution:
<em>This problem is related to projectile motion for objec</em>t
First calculate the Range for object and it is given by ,
(2Ф)/
Here R= range g= acceleration due to gravity =9.8 m/sec^2
1m =3.2 feet
So 9.8 m, equals to 9.8 *3.2=31.36 ft
So g=31.36 ft/sec^2. and 2Ф=2(28)=56


fts
Now using Formula for time and range as

Vx is horizontal velocity
Ф
(28)
ft/sec
So above equation becomes as ,


T is approximately equals to 4 sec.
Answer:
Step-by-step explanation:
We will use 2 coordinates from the table along with the standard form for an exponential function to write the equation that models that data. The standard form for an exponential function is
where x and y are coordinates from the table, a is the initial value, and b is the growth/decay rate. I will use the first 2 coordinates from the table: (0, 3) and (1, 1.5)
Solving first for a:
. Sine anything in the world raised to a power of 0 is 1, we can determine that
a = 3. Using that value along with the x and y from the second coordinate I chose, I can then solve for b:
. Since b to the first is just b:
1.5 = 3b so
b = .5
Filling in our model:

Since the value for b is greater than 0 but less than 1 (in other words a fraction smaller than 1), this table represents a decay function.
Answer:68.2
Step-by-step explanation:
cos6=71/EF
cos6= 0.960170286650366
0.960170286650366= 71/EF
0.960170286650366*71=EF
EF= 68.2