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vfiekz [6]
4 years ago
8

At a yearly basketball tournament, 84 different teams complete, after each round of the tournament, one-fourth of the team remai

n.
Create a function which models the above scenario and explain how you arrived at each part of the function
Mathematics
1 answer:
Mariana [72]4 years ago
6 0

Answer:

84*(1/4)^x

Step-by-step explanation:

84 is the constant as it is the starting amount. However each round (x because the amount of rounds is constantly changing) divides the team by 4 or multiplies by 1/4 (Reciprocal).

So, if you wanted to see if the amount of teams after three rounds you would do:

84 * 1/4 * 1/4 * 1/4. However, any number times itself is an exponent, so we would simplify to 84³. But due to the fact we do not know how many rounds we are going to calculate we change the exponent to x.

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3 years ago
What does the y-intercept of the graph of the function f(m) represent? (2 points)
Marina86 [1]

Answer:

you do realize you're showing your face

Step-by-step explanation:

4 0
3 years ago
A cylinder has a diameter of 6 ft and height of 9 ft. What is the volume of this cylinder? A) 54 ft2 B) 169.8 ft3 C) 254.3 ft3 D
Margarita [4]
<span>In order to find the volume of a cylinder, we must find the area of its circular face and multiply that value by its height. The area of a circle is (pi)*r^2. The diameter is twice the radius, so the cylinder has a radius of 6/2, or 3 ft. The area of the circular face of the cylinder is therefore 9(pi) or approximately 28.27 sq. ft. Multiplying that area by the height of 9 ft. results in a volume of approximately 254.3 cubic ft., which is answer C.</span>
7 0
3 years ago
Read 2 more answers
Which one would be correct
Oduvanchick [21]

Answer:

x=-35

Step-by-step explanation:

8 0
3 years ago
Evaluate the infinite sum:
satela [25.4K]

Consider the <em>k</em>-th partial sum,

S_k = 1 + \dfrac2\pi + \dfrac3{\pi^2} + \cdots + \dfrac k{\pi^{k-1}}

More compactly,

\displaystyle S_k = \sum_{i=1}^k \frac i{\pi^{i-1}} = \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}}

(this is just another case of a similar sum you asked about a while ago [24494877])

The infinite sum is the limit of the partial sum as <em>k</em> goes to infinity. We have

\displaystyle \lim_{k\to\infty} \frac{(1-\pi)k+\pi^{k+1}-\pi}{(1-\pi)^2\pi^{k-1}} = \frac\pi{(1-\pi)^2} \lim_{k\to\infty} \left(\frac{(1-\pi)k}{\pi^k} + \pi - \frac1{\pi^{k-1}} \right) = \boxed{\frac{\pi^2}{(1-\pi)^2}}

since the non-constant terms in the limit converge to 0.

Alternatively, recall that for |<em>x</em>| < 1, we have

\dfrac1{1-x} = \displaystyle \sum_{n=0}^\infty x^n

Differentiating both sides gives

\dfrac1{(1-x)^2} = \displaystyle \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}

also valid for |<em>x</em>| < 1. Take <em>x</em> = 1/<em>π</em> and you get the sum you want to compute.

5 0
3 years ago
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