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IgorC [24]
2 years ago
15

Usian bolt can run 100meters in 9.58 seconds . my car can drive 100kilometers an hour. which is faster, is it faster to travel 1

00meters in 9.58 seconds or 100 kilometers an hour
Mathematics
1 answer:
Mashcka [7]2 years ago
5 0

Answer:

Your car is faster of course

Step-by-step explanation:

100 km would be 100000 meters. Then you would need to know that if Usain bolt ran 100 meters every 9.54 sec then in 100000 meters he would have ran 9540 seconds which is also known has 159 minutes, which is more than 1 hour so your car is faster.

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The diameter of the circle is 9.7 m . Calculate the length of its circumference. Round to the nearest tenth and remember to use
RSB [31]

Answer:

30.4m

Step-by-step explanation:

Diameter(D) = 2r (radius)

D = 9.7m

r = 9.7m / 2

= 4.85m

C = 2πr

= 2π × 4.85m

= 30.47344873982099m

rounding - Leave it as is if the digit in the tenth is 4 or lower

- round it up if it's 5 or higher.

~~ 30.4m

5 0
2 years ago
Find the mass of the solid paraboloid Dequals=​{(r,thetaθ​,z): 0less than or equals≤zless than or equals≤8181minus−r2​, 0less th
Lubov Fominskaja [6]

Answer:

M = 5742π  

Step-by-step explanation:

Given:-

- Find the mass of a solid with the density ( ρ ):

                             ρ ( r, θ , z ) = 1 + z / 81

- The solid is bounded by the planes:

                             0 ≤ z ≤ 81 - r^2

                             0 ≤ r ≤ 9

Find:-

Find the mass of the solid paraboloid

Solution:-

- The mass (M) of any solid body is given by the following triple integral formulation:

                           M = \int \int \int {p ( r ,theta, z)} \, dV\\\\

- We can write the above expression in cylindrical coordinates:

                           M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\

- Perform integration:

                           M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\

                           M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] }  \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] }  \, dtheta\\\\M = 5742\pi  kg              

- The mass evaluated is M = 5742π                      

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Answer:

In problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. To find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

Step-by-step explanation:

In problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. To find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

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