H=+15 m
v=+5 m/s
Ball hits the ground when h(t)=-15m
h(t)=-9.8t^2+vt+h
=>
-15=-9.8t^2+5t+15
9.8t^2-5t-30=0
Solve for t, using quadratic formula,
t=-1.513 or t=2.023
reject negative root due to context, so
t=2.023 seconds
2)
h(t)=-16t^2+20t+8
a. height before pitch is when t=0, or h(0)=8
b. highest point reached when h'(t)=-32t+20=0 => t=5/8 seconds
c. highest point is t(5/8)=-16(5/8)^2+20(5/8)+8=47/5=9.4 m
d. ball hits ground when h(t)=0 => solve t for h(t)=0
=> t=-0.3187 seconds or t=1.569 seconds.
Reject negative root to give
time to hit ground = 1.569 since ball was pitched.
Answer:
12 and 16 or 12:16
6 x 2 = 12
8 x 2 =16
12 + 16 = 28
Step-by-step explanation:
Let's begin ! ^_^
You have got informations in your problem that you just have to translate in a "mathematic language".
Company A :
"$40 membership fee and $2 per video stream."
Thanks to that, you can guess :
c = 40 + 2s
Now let's do the same thing to the company B.
Company B :
"charges a one-time $20 membership fee and $4 per video stream."
Therefore, c = 20 + 4s.
I think you have guessed !! =D And yes the system of equation that you have to solve and you're looking for is :
{c=40+2s ( Option B)
c=20+4s.
Second step : For how many video streams will the cost be the same for both companies?
We just have to solve the system :) And there is only one variable, the letter "s".
We know that we have to find equal costs as it is said in the question " the cost be the same for both companies", so :
=> 40 + 2s = 20 +4s
=> 4s - 2s = 40 - 20
=> 2s = 20
=> s = 20/2
=> s = 10
Verification :
Company A :
=> 40 + 2s = 40 + 2*10 = 40 + 20 =60.
Company B :
=> 20 + 4s = 20 + 4*10 = 20 + 40 = 60
As you have seen, the costs are the same for both companies.
We can say that as a conclusion for 10 video streams, the cost will be the same for both companies.
In short, the answer would be : 10.
Let me give you an advice : Usually the word "per" means a multiplication is the mathematic language.
Hope this helps !
Photon
We have that
x²<span> - y = -2
4y - 8 = x
using a graph tool
see the attached figure
the answer in the attached figure</span>
