Answer:
R ≈ 8.31
Step-by-step explanation:
Filling in the given values ...
... pV = nRT
... 2·4.155 = 1·R·1
... 8.31 = R
300000+70000+3000+600+90+8
9514 1404 393
Answer:
Step-by-step explanation:
A graphing calculator answers these questions easily.
The ball achieves a maximum height of 40 ft, 1 second after it is thrown.
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The equation is usefully put into vertex form, as the vertex is the answer to the questions asked.
h(t) = -16(t^2 -2t) +24
h(t) = -16(t^2 -2t +1) +24 +16 . . . . . . complete the square
h(t) = -16(t -1)^2 +40 . . . . . . . . . vertex form
Compare this to the vertex form:
f(x) = a(x -h)^2 +k . . . . . . vertex (h, k); vertical stretch factor 'a'
We see the vertex of our height equation is ...
(h, k) = (1, 40)
The ball reaches a maximum height of 40 feet at t = 1 second after it is thrown.
5w^2 = A
You would multiply the width (w) by 5 because it's 5 times more than twice it's width. Then w (Width) would be to the power of 2 because it's 5 times more than twice its width. And this will all equal the Area (a)
Answer:
at t=0 the car has moved 0 feet, also you know it has a constant speed so you get the points:
P1:(0,0) therefore you know b, the y-intercept=0
P2:(8,840)
insert P2 into y=mx+b
840=8m+0
840/8=m
105=m
so the result is:
y=105x=105t=f(t)=d
Step-by-step explanation: