What is h = -16a² + 1700 (function)

1 answer:

3 0

I got
t = 6.614 -16t^2 +1700 = 1000 -16t^2 = -700 t^2 = 43.75
You need to divide by 16 first, before you take the square root.
Domain = what values of time would be used here? t > 0, since negative time not reasonable. And you probably need to find when the ball hits the ground, since after that time, the ball would be underground.
So the a reasonable domain is from 0 to 10.3078
Range: what values of height seem reasonable? h > 0, because negative height would mean that the ball was underground. And ... since the ball was "dropped" from a height of 1700 (vs being "thrown up" from a height) the maximum height would be 1700. So ... a reasonable range would be from 0 to 1700.

Hope this helps... :)

You might be interested in

roger completed a probability experiment with a coin. he flipped the coin 32 times, and it landed on tails eight times. he looke

vladimir2022 [97]

I don't know what this question is asking exactly, but if i am right, it is asking for the ratio of heads flipped to tails flipped. If this is correct, then the ratio would be 24 to 8. My reasoning for this is, it already tells you that you have 8 times that tails was flipped, and all you have to do to figure out the amount of times heads if flipped it to subtract: 32-8=24, so that means that heads was flipped a total of 24 times. The last thing that you have to do is to put it into a ratio, and in the order it gives you initially, which would make the answer (3 ways):

1) 24 to 8
2) 24 : 8
3) 24/8

I hope this helps you.

4 0

4 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$