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2 years ago

There are nine players on a baseball team or club has nine baseball teams how many baseball players are in the club

Mathematics

2 answers:

ohaa [14]2 years ago

8 0

Answer:

The club has 81 players

Step-by-step explanation:

If each team has 9 players and the club has 9 teams all you have to do is multiply 9*9

tankabanditka [31]2 years ago

3 0

Answer: 81

Step-by-step explanation:

We are given that ,

The number of players in a baseball team = 9

The number of baseball teams in a club = 9

To find : The total number of baseball players are in the club.

The total number of baseball players are in the club = (Number of players in a baseball team) x (Number of baseball teams in a club)

= 9 x 9

= 81

Hence, the total number of baseball players are in the club is 81.

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Blababa [14]

Answer:

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Step-by-step explanation:

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So, putting the value

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2 years ago

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V125BC [204]

Answer:

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1 year ago

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V125BC [204]

Answer:

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Step-by-step explanation:

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2 years ago

The graph of a function f is shown. Use the graph to estimate the average rate of change from x = –2 to x = 0

vodomira [7]

Answer:

Average rate of change = 2

Step-by-step explanation:

Formula to get the rate of change of a function between x = b and x = a is,

Rate of change = $\frac{f(b)-f(a)}{b-a}$

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From the graph attached,

At x = -2,

f(-2) = 0

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Average rate of change of the given function between x = -2 and x = 0 is 2.

4 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%20%5Crm%20%20%5Clim_%7Bk%20%5Cto%20%5Cinfty%20%7D%20%5Csqrt%5B%20%20k%5D%7B%20%5CGamma%20%

Naddik [55]

We have

$\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)$

and as k goes to ∞, the exponent converges to a definite integral. So the limit is

$\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}$

6 0

1 year ago