I'm quite confused with the racetrack principle and need on this problem. . suppose that f(t) is continuous and twice-differentiable for t> = 0. further suppose f''(t) > = 3 for all t> = 0 and f(0) = f'(0) = 0.. using the racetrack principle, what linear function g(t) can we prove is less than f'(t) (for t> = 0)? . g(t) = . then, also using the racetrack principle, what quadratic function h(t) can we prove is less than than f(t) (for t> = 0)? . h(t) = .

Race track principle says that if two functions are equal at t=0, then the one which has a greater derivative will be greater.

In this case we're comparing f′(t) and g′(t). So we make sure that g(0)=f′(0) and that f′′(t)≥g′(t)

g(t)=at+b

Since it is a line.
g′(t)=a

f′′(t)≥3≥g′(t)⟹3≥a

So let a=3.

So

h(t)=at2+bt

the largest number of the four options is 1.8 x 10-2

you have it in scientific notation.

1.80 x 10-6   = 0.00000180

1.80 x 10-4   = 0.000180 (bigger, or "less small", than the former one)

so

1.8 x 10-2       = 0.0180. as you see. the "less small", or the biggest one of them all.

as you have negative exponent, it means you are dividing 1.8 by 100 (i.e. moving the dot towards left, two times).

if you had positive exponent for the scientific notation, power of tenths, you would be multiplying times 100 (moving the dot towards right) as below:

1.8 x (10)2 = 180

so. if you have negative exponent in scientific notation, move the dot toward left. if it is positive, toward right as many times the exponent tells you.

step-by-step explanation: given equation a = h(b 1+b 2)

also we are given values of a, h and b 1 as :

a = 16, h = 4, and b 1 = 3.

in order to solve it for b 2, we need to plug the given values of   a, h and b 1 .

therefore, plugging values of   a, h and b 1   in given equation, we get

16 = 4( 3 + b 2)

first dividing both sides by 4, we get

4 = 3 + b 2

subtracting 3 from both sides, we get

4 -3 = 3 -3 + b 2

1 = b 2.