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ladessa [460]
2 years ago
8

Three teams, A, B and C, play in a competition.

Mathematics
1 answer:
MatroZZZ [7]2 years ago
5 0

Answer:

<u>47.5 games</u>

Step-by-step explanation:

Finding A :

⇒ Games by A : Games by B = 3 : 1

⇒ Games by A : 10 = 3 : 1

⇒ Games by A = 30

==============================================================

Finding C :

⇒ Games by B / Games by C × Games by A / Games by B = 3 × 4/3

⇒ Games by A / Games by C = 4

⇒ Games by C = 30/4

⇒ Games by C = 7.5

==============================================================

Total games won :

⇒ Games by A + Games by B + Games by C

⇒ 30 + 10 + 7.5

⇒ <u>47.5 games</u>

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Step-by-step explanation:

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3 years ago
Two similar cones have volume of 4 m and 108 m respectively. If the large one has surface area 54 m, find the surface area of th
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\displaystyle\bf\\Explanations:\\\\The~similarity~ratio~of~two~similar~cones=k\\\\k=is~the~ratio~between~2~corresponding~lengths\\\\k=\frac{R_1}{R_2}=\frac{h_1}{h_2}\\\\The~ratio~between~2~corresponding~areas~of~similar~cones=k^2\\\\\frac{Area~1}{Area~2}=k^2\\\\The~ratio~between~the~volumes~of~the~2~similar~cones=k^3\\\\\frac{Volume~1}{Volume~2}=k^3

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\displaystyle\bf\\Solving:\\\\Volume1=108~m^3\\\\Volume2=4m^3\\\\k^3=\frac{108}{4}\\\\k^3=27~~\Big|\sqrt{~}\\\\k=\sqrt[\b3]{27}\\\\k=3\\\\Area1=54~m^2\\\\\frac{Area~1}{Area~2}=k^2\\\\\frac{54}{Area~2}=3^2\\\\\frac{54}{Area~2}=9\\\\Area2=\frac{54}{9}\\\\\boxed{\bf Area2=6 m^2}

 

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3 years ago
Which is the side length of a cube with a volume of 3375 in3?
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Newton’s law of cooling states that dx/dt= −k(x − A) where x is the temperature, t is time, A is the ambient temperature, and k
Vadim26 [7]

Answer:

(a) The solution to the differential equation is x = A_0Coswt + Ce^(-kx)

(b) The initial condition t > 0 will not make much of a difference.

Step-by-step explanation:

Given the differential equation

dx/dt= −k(x − A); t > 0, A = A_0Coswt

(a) To solve the differential equation, first separate the variables.

dx/(x - A) = -kdt

Integrating both sides, we have

ln(x - A) = -kt + c

x - A = Ce^(-kt) (Where C = ce^(-kt))

x = A + Ce^(-kx)

Now, we put A = A_0Coswt

x = A_0Coswt + Ce^(-kx) (Where C is constant.)

And we have the solution.

(b) Since temperature t ≠ 0, the initial condition t > 0 will not make much of a difference because, Cos(wt) = Cos(-wt).

It is not any different from when t < 0.

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3 years ago
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