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Helga [31]
2 years ago
12

Find the length of side xx in simplest radical form with a rational denominator.

Mathematics
1 answer:
anygoal [31]2 years ago
8 0

Answer:

x =   \frac{3}{ \sqrt{2} }

Step-by-step explanation:

{x}^{2}  +  {x}^{2}  =  {3}^{2}  \\ 2 {x}^{2}  = 9 \\  {x}^{2}  =  \frac{9}{2}  \\ x =  \frac{3}{ \sqrt{2} }  \\ x = 2.121

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You and five friends have joined baseball.The team membership fee is $50 plus a $5 fee per game to pay the referee.
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Which set of ordered pairs represents a function?
Mrac [35]
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2 years ago
(1 point) Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference betw
slavikrds [6]

Answer:

a. \frac{dT}{dt}=k(T-Tm); T(0)=190

b. C_{0}=122

c. k=-0.00259

d. t=153.39838\\ minutos

Step-by-step explanation:

a. Newton's law of cooling states that the speed with which a body is cooled is proportional to the difference between its temperature and that of the medium in which it is found. Then, the initial value problem is given by:

Tm=68

\frac{dT}{dt}=k(T-Tm); T(0)=190

b. The differential equation obtained is a differential equation of separable variables:

\frac{dT}{T-Tm}=kdt\\\\\int {\frac{dT}{T-Tm}}=\int{kdt}\\\\Ln|T-Tm|=kt+C\\\\T(t)=C_{0}e^{kt}+Tm=C_{0}e^{kt}+68\\\\T(0)=C_{0}e^{k(0)}+68=190\\\\C_{0}=122

c. After 33 minutes of serving the coffee has cooled to 180°:

T(33)=122e^{33k}+68=180\\\\e^{33k}=\frac{112}{122}\\\\33k=Ln(\frac{112}{122})\\\\k=-0.00259

d.

150=122e^{-0.00259t}+68\\\\Ln(\frac{150-68}{122})=-0.00259t\\\\t=153.39838\\\\

8 0
3 years ago
What can you tell about the mean of eachdistribution
erik [133]

Answer:

In a discrete probability distribution of a random variable X, the mean is equal to the sum over every possible value weighted by the probability of that value; that is, it is computed by taking the product of each possible value x of X and its probability p(x), and then adding all these products together, giving. .

Step-by-step explanation:

I really hope this helps have a wonderful day

7 0
3 years ago
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