Answer:
r ≥ 6
Step-by-step explanation:
37+15r ≥ 127
-37 -37 37 cancels and you're left with 15r
15r ≥ 90
/ 15 /15 15 cancels and you're left with r
r ≥ 6
The scale factor of the smaller figure to the larger figure is 4/25
<h3>Scale factors</h3>
Scale factors is used to enlarge or diminish the size of an object. The rato of the sides or volume of the figures is the scale factor.
Given the following
Surface area of small cylinder = 20πyd²
Surface area of large cylinder = 125πyd²
find the ratio
Scale factor = 20π/125π
Scale factor = 4/25
Hence the scale factor of the smaller figure to the larger figure is 4/25
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Answer:
The point P' is the image of the point P under the translation
The coordinates of the point P are (6,0)
We need to determine the coordinates of the point P'
Coordinates of the point P':
The coordinates of the point P' can be determined by substituting the coordinates of the point P(6,0) in the translation.
Thus, substituting the coordinates, we have;
Simplifying the coordinates, we get;
Thus, the coordinates of the point P' is (0,-1)
Step-by-step explanation:
1/4(6) + 3(7)
6/4 + 21
3/2 + 42/2
45/2= 22 1/2
a. The velocity t = ![v = Ce_{n} (\frac{mo}{mo - kt} ) - gt](https://tex.z-dn.net/?f=v%20%3D%20Ce_%7Bn%7D%20%28%5Cfrac%7Bmo%7D%7Bmo%20-%20kt%7D%20%29%20-%20gt)
b. v60 = 7164
<h3>How to solve for the velocity</h3>
mdv/dt = ck - mg
dv/dt = ck/m - mg/m
= ck/m - g
dv = ![(\frac{ck}{Mo-Kt} -g)dv](https://tex.z-dn.net/?f=%28%5Cfrac%7Bck%7D%7BMo-Kt%7D%20-g%29dv)
Integrate the two sides of the equation to get
v ![-\frac{ck}{k} e_{n} (Mo- kt)-gt+c](https://tex.z-dn.net/?f=-%5Cfrac%7Bck%7D%7Bk%7D%20e_%7Bn%7D%20%28Mo-%20kt%29-gt%2Bc)
![v = Ce_{n} (\frac{mo}{mo - kt} ) - gt](https://tex.z-dn.net/?f=v%20%3D%20Ce_%7Bn%7D%20%28%5Cfrac%7Bmo%7D%7Bmo%20-%20kt%7D%20%29%20-%20gt)
b. fuel accounts for 55% of the mass
So final mass after fuel is burned out is = 0.45
c=2500
g=9.8
t=60
v = -2500ln0.45 - 9.8 x 60
= 7752 - 588
= 7164
<h3>Complete question</h3>
A rocket, fired from rest at time t = 0, has an initial mass of m0 (including its fuel). Assuming that the fuel is consumed at a constant rate k, the mass m of the rocket, while fuel is being burned, will be given by m0 - kt. It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed c relative to the rocket, then the velocity of the rocket will satisfy the equation where g is the acceleration due to gravity.
dv dt m =ck - mg
(a) Find v(t) keeping in mind that the mass m is a function of t.
v(t) =
m/sec
(b) Suppose that the fuel accounts for 55% of the initial mass of the rocket and that all of the fuel is consumed at 60 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take g = 9.8 m/s² and c = 2500 m/s.]
v(60) =
m/sec [Round to nearest whole number]
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The answer would be 288 to the nearest whole number