80 degrees because it decreases by a half each time so by the time it is E (the fourth letter) it has gotten to 20 so the answer is 160
5 × 1. = 5
_ × _. = _
8 × 4. = 32
5/32
Plug the value -5 in. (3.4)(-5) - 8 is equivalent to -17 - 8. Final answer: g(-5) = -25.
Answer:
The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction
Step-by-step explanation:
- For a constant acceleration:
, where
is the final velocity in a direction after the acceleration is applied,
is the initial velocity in that direction before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied. - <em>Then for the x direction</em> it is known that the initial velocity is
5320 m/s, the acceleration (the applied by the engine) in x direction is
1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the is 739 s. Then: 
- In the same fashion, <em>for the y direction</em>, the initial velocity is
0 m/s, the acceleration in y direction is
7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: 
<h2>
Answer with explanation:</h2>
It is given that:
f: R → R is a continuous function such that:
∀ x,y ∈ R
Now, let us assume f(1)=k
Also,
( Since,
f(0)=f(0+0)
i.e.
f(0)=f(0)+f(0)
By using property (1)
Also,
f(0)=2f(0)
i.e.
2f(0)-f(0)=0
i.e.
f(0)=0 )
Also,
i.e.
f(2)=f(1)+f(1) ( By using property (1) )
i.e.
f(2)=2f(1)
i.e.
f(2)=2k
f(m)=f(1+1+1+...+1)
i.e.
f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)
i.e.
f(m)=mf(1)
i.e.
f(m)=mk
Now,

Also,
i.e. 
Then,

(
Now, as we know that:
Q is dense in R.
so Э x∈ Q' such that Э a seq
belonging to Q such that:
)
Now, we know that: Q'=R
This means that:
Э α ∈ R
such that Э sequence
such that:

and


( since
belongs to Q )
Let f is continuous at x=α
This means that:

This means that:

This means that:
f(x)=kx for every x∈ R