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Helga [31]
3 years ago
7

A graph titled Swimmers at City Pool has Temperature (degrees Fahrenheit) on the x-axis, and swimmers on the y-axis. A trend lin

e goes through points (75, 25) and (85, 40).
Compare the number of swimmers at City Pool and the outside temperature to make an interpolation.

What is the best estimate for the number of swimmers when the outside temperature is 75ºF?
15 swimmers
20 swimmers
25 swimmers
30 swimmers
Mathematics
2 answers:
aniked [119]3 years ago
7 0

Answer:

25 is the answer

skelet666 [1.2K]3 years ago
3 0

Answer:

25

Step-by-step explanation:

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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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rusak2 [61]

Answer:

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4x - 5(x + 3)         Remove the brackets. Watch the - sign and -5*3

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What is 8% of 75?.?.?
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Step-by-step explanation:

(8:100) * 75

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