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Norma-Jean [14]

3 years ago

7

andre,rick,and ty were practicing for a track cycling race. andre covered a distance of 750m in 3 min. rick clocked 4 minutees f

or a distace of 800 m. ty raced 950 m in 5 minutes. who coveredthe least distance per minute

2 answers:

FrozenT [24]3 years ago

8 0

Answer:

Ty covered the least distance per min.

Step-by-step explanation:

Who covered the shortest distance per minute, or who rode his bike most slowly?

Andre's rate: (750 m) / (3 min) = 250/3 m/min

Rick's rate: (800 m) / (4 min) = 200 m/min

Ty's rate: (950 m) / (5 min) = 190 m/min

Ty was the rider who covered the least distance per minute; his rate was the slowest.

Ilya [14]3 years ago

6 0

Answer: Ty covered the least distance per minute

Step-by-step explanation:

Hi, to answer this question we have to calculate the speed ratio for each case, using the formula:

<em>Speed = distance (meters) / time (minutes) </em>

Andre = 750/3 =250 meters per minute
Rick = 800/4 =200 meters per minute
Ty = 950/5 = 190 meters per minute

Analyzing the results, Ty covered the least distance per minute because Ty has the lowest speed rate.

190 <200<250

Ty<Rick<André

Feel free to ask for more if needed or if you did not understand something.

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Step-by-step explanation:

Since, we know that,

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Where,

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h = height,

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Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

$4^2 = (2r)^2 + h^2$ ( see in the below diagram ),

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$\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})$

$=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})$

$=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})$

Again differentiating with respect to r,

$\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})$

For maximum or minimum,

$\frac{dS}{dt}=0$

$2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0$

$-8r^2 + 16 = 0$

$8r^2 = 16$

$r^2 = 2$

$\implies r = \sqrt{2}$

Since, for r = √2,

$\frac{d^2S}{dt^2}=negative$

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

$S = 2\pi (\sqrt{2})(\sqrt{16-8})$

$=2\pi (\sqrt{2})(\sqrt{8})$

$=2\pi \sqrt{16}$

$=8\pi\text{ square cm}$

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