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3 years ago

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A soccer team has 11 players on the field. Of those players, 2 are forwards, 4 are midfielders, 4 are defenders, and 1 is a goal

ie. 11 players on a soccer field. What is the fraction of players on the field that are midfielders?

1 answer:

Murrr4er [49]3 years ago

3 0

Answer:

4/11

Step-by-step explanation:

There are 4 midfielders on the team out of 11.

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The height h(n) of a bouncing ball is an exponential function of the number n of bounces.

Digiron [165]

Answer:

The height of a bouncing ball is defined by $h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}$ .

Step-by-step explanation:

According to this statement, we need to derive the expression of the height of a bouncing ball, that is, a function of the number of bounces. The exponential expression of the bouncing ball is of the form:

$h = h_{o}\cdot r^{n-1}$ , $n \in \mathbb{N}$ , $0 < r < 1$ (1)

Where:

$h_{o}$ - Height reached by the ball on the first bounce, measured in feet.

$r$ - Decrease rate, no unit.

$n$ - Number of bounces, no unit.

$h$ - Height reached by the ball on the n-th bounce, measured in feet.

The decrease rate is the ratio between heights of two consecutive bounces, that is:

$r = \frac{h_{1}}{h_{o}}$ (2)

Where $h_{1}$ is the height reached by the ball on the second bounce, measured in feet.

If we know that $h_{o} = 6\,ft$ and $h_{1} = 4\,ft$ , then the expression for the height of the bouncing ball is:

$h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}$

The height of a bouncing ball is defined by $h(n) = 6\cdot \left(\frac{4}{6} \right)^{n-1}$ .

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3 years ago

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schepotkina [342]

Answer:

2/6

Step-by-step explanation:

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2 years ago

Read 2 more answers

Becky accepted a job as an inventory manager after being offered an $800 signing bonus. If she makes $23 an hour, which equation

dusya [7]

Y = 23x + 800 (Apex)

6 0

3 years ago

Let P (n) be the statement that a postage of n cents can be formed using just 4-cent stamps and 7-cent stamps. The goal is to sh

natulia [17]

Complete Question

Let P (n) be the statement that a postage of n cents can be formed using just 4-cent stamps and 7-cent stamps. The parts of this exercise outline a strong induction proof that P (n) is true for n ≥ 18.

Show statements P (18), P (19), P (20), and P (21) are true, completing the basis step of the proof.

Answer:

P(18) is true

P(19) is true

P(20) is true

P(21) is true

Step-by-step explanation:

a. When n = 18

18 cents can be formed using two 7cents and one 4cents

i.e. 2 * 7 + 4 = 18

So, P(18) is true

b. When n = 19

19 cents can be formed using one 7cents and three 4cents

i.e. 1 * 7 + 3 * 4 = 19

So, P(19) is true

c. When n = 20

18 cents can be formed using five 4cents

i.e. 5 * 4 = 20

So, P(20) is true

d. When n = 21

18 cents can be formed using three 7cents

i.e. 3 * 7 = 21

So, P(21) is true

7 0

3 years ago

ACTIVITY 2 (19) Mr Duma recently inherited a rectangular plot, part of the estate left by his late father. The plot with the fol

seraphim [82]

The sum of the lengths of three sides of the rectangle, gives the length

of the fencing, while one third of the rectangle area is for the pavement.

Responses:

Project A: The formula for the length of the fencing is, L = 4·x - 1

Project B: Length of the fancy wall = 2·x + 1

$Project \ C:Area \ of \ the \ paving = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$

<h3>Which method can be used to find the length and area of paving from the given equations?</h3>

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Given parameters are;

Length of the rectangular plot = 2·x + 1

Width of the rectangular plot = x - 1

Vertices of the rectangular plot are; QPSR

Project A: Let QP and SR represent the longest sides of the rectangle, we have;

PQ = SR = 2·x + 1

Which gives;

SP = QR = x - 1

The length of the fencing, L = SP + PQ + QR = x - 1 + 2·x + 1 + x - 1 = 4·x - 1

The formula for the length of the fencing is, L =<u> 4·x - 1</u>

Project B: The front side is SR

Therefore;

Length of the fancy wall = <u>2·x + 1</u>

Project C:

Area, A = Length × Width

Area of the plot, A = (2·x + 1) × (x - 1) = 2·x² - x - 1

$Area \ of \ the \ paving = \dfrac{1}{3} \times \left(2 \cdot x^2 - x - 1\right) = \underline{ \dfrac{2\cdot x^2 }{3} - \dfrac{x }{3} - \dfrac{1}{3}}$

Learn more about writing equations here:

brainly.com/question/24760633

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2 years ago