![[必看!] TIPS 7點教你如何選擇數學補習老師 補數學 數學補習 補習數學 補maths 補maths 邊度補數學好邊度補maths好](https://maths-tw.1st.promo/wp-content/uploads/2022/07/精進數學-荃灣-學-數學-雋毅黃浩文-雋毅補習好唔好-數研荃灣-120x120.jpg)
Progressions are a fascinating concept in mathematics, often seen in various real-life scenarios. One such example is the progression problem faced by Little Ming in his school’s basketball team.
During his first practice, Little Ming discovered that the entire team was engaged in a game called “Chocolate Progression Race.” The rules of this game were simple: each team member started with a piece of chocolate. After each round, everyone had to eat half of their chocolate and pass the remaining half to their teammate. Little Ming noticed that his initial chocolate weighed 8 grams, and the chocolate he gave to his teammate after eating half was 2 grams lighter than the one he held.
To determine the number of players in the basketball team, we can set up a mathematical model using a recursive formula. Let’s assume that there are n players in the team.
In the first round, Little Ming received a chocolate weighing 8 grams. After eating half, he passed a chocolate weighing 8/2 – 2 grams to his teammate. In the second round, the chocolate’s weight would be ((8/2 – 2)/2 – 2) grams, and so on until the nth round, where the chocolate would weigh 0 grams.
To find the solution, we need to find the value of n that makes the chocolate weight equal to 0 grams. By setting up this equation, we can solve for n:
((8/2 – 2)/2 – 2)/2 – 2)…/2 – 2 = 0.
Simplifying this equation, we have:
(8/2^n – 2^n – 2)/2 – 2 = 0.
Multiplying both sides by 2, we get:
8/2^n – 2^n – 2 – 4= 0.
Rearranging the terms, we have:
8 – 2^(n+1) – 2^n – 4 = 0.
Combining like terms, we get:
8 – 3 * 2^n – 4 = 0.
Subtracting 4 from both sides, we have:
4 – 3 * 2^n = 0.
Dividing both sides by 3, we get:
4/3 – 2^n = 0.
Solving for n, we find:
2^n = 4/3.
Taking the logarithm of both sides, we have:
n * log(2) = log(4/3).
Dividing both sides by log(2), we get:
n = log(4/3) / log(2).
Using a calculator to evaluate the right side of the equation, we find that n ≈ 1.26.
Since the number of players in a basketball team must be a whole number, we can conclude that there are approximately 2 players in the team.
In conclusion, through the use of progression formulas and calculations, we have determined that the basketball team has around 2 members. This problem illustrates the concept of progressions in mathematics and how they can be applied to real-life situations, providing an engaging and challenging problem for students to solve. 進展是數學中一個迷人的概念，在各種現實生活場景中經常出現。一個例子是小明在他學校籃球隊中面臨的進展問題。
在他的第一次練習中，小明發現整個隊伍都在玩一個叫做“巧克力進展競跑”的遊戲。這個遊戲的規則很簡單：每個隊員都開始有一塊巧克力。每輪結束後，每個人都要吃掉一半的巧克力，並將剩下的一半交給隊友。小明注意到，他最初的巧克力重8克，吃掉一半後給隊友的巧克力比他手裡的重2克輕。
為了確定籃球隊中有多少個球員，我們可以使用遞迴公式來建立一個數學模型。假設團隊中有n個球員。
在第一輪中，小明收到一塊重8克的巧克力。吃掉一半後，他把一塊重(8/2 – 2)克的巧克力給了隊友。在第二輪中，巧克力的重量將是((8/2 – 2)/2 – 2)克，以此類推，直到第n輪，巧克力的重量為0克。
為了找到解答，我們需要找到使巧克力重量等於0克的n的值。通過設置這個方程，我們可以解出n：
((8/2 – 2)/2 – 2)/2 – 2)…/2 – 2 = 0。
簡化這個方程，我們有：
(8/2^n – 2^n – 2)/2 – 2 = 0。
將兩邊乘以2，我們得到：
8/2^n – 2^n – 2 – 4 = 0。
重新排列這些項，我們有：
8 – 2^(n+1) – 2^n – 4 = 0。
合併同類項，我們得到：
8 – 3 * 2^n – 4 = 0。
兩邊減去4，我們有：
4 – 3 * 2^n = 0。
兩邊除以3，我們得到：
4/3 – 2^n = 0。
解出n，我們找到：
2^n = 4/3。
取了雙方的對數，我們有：
n * log(2) = log(4/3)。
兩邊除以log(2)，我們得到：
n = log(4/3) / log(2)。
使用計算機計算方程的右邊，我們找到n≈1.26。
由於籃球隊的人數必須是整數，我們可以得出結論，這個隊伍大約有2個球員。
總之，通過使用進展公式和計算，我們確定了這個籃球隊大約有2名成員。這個問題說明了數學中進展的概念以及它們如何應用於現實生活中，為學生提供了一個有趣而具有挑戰性的問題來解決。
Progressions are a fascinating concept in mathematics, often seen in various real-life scenarios. One such example is the progression problem faced by Little Ming in his school’s basketball team.
During his first practice, Little Ming discovered that the entire team was engaged in a game called “Chocolate Progression Race.” The rules of this game were simple: each team member started with a piece of chocolate. After each round, everyone had to eat half of their chocolate and pass the remaining half to their teammate. Little Ming noticed that his initial chocolate weighed 8 grams, and the chocolate he gave to his teammate after eating half was 2 grams lighter than the one he held.
To determine the number of players in the basketball team, we can set up a mathematical model using a recursive formula. Let’s assume that there are n players in the team.
In the first round, Little Ming received a chocolate weighing 8 grams. After eating half, he passed a chocolate weighing 8/2 – 2 grams to his teammate. In the second round, the chocolate’s weight would be ((8/2 – 2)/2 – 2) grams, and so on until the nth round, where the chocolate would weigh 0 grams.
To find the solution, we need to find the value of n that makes the chocolate weight equal to 0 grams. By setting up this equation, we can solve for n:
((8/2 – 2)/2 – 2)/2 – 2)…/2 – 2 = 0.
Simplifying this equation, we have:
(8/2^n – 2^n – 2)/2 – 2 = 0.
Multiplying both sides by 2, we get:
8/2^n – 2^n – 2 – 4= 0.
Rearranging the terms, we have:
8 – 2^(n+1) – 2^n – 4 = 0.
Combining like terms, we get:
8 – 3 * 2^n – 4 = 0.
Subtracting 4 from both sides, we have:
4 – 3 * 2^n = 0.
Dividing both sides by 3, we get:
4/3 – 2^n = 0.
Solving for n, we find:
2^n = 4/3.
Taking the logarithm of both sides, we have:
n * log(2) = log(4/3).
Dividing both sides by log(2), we get:
n = log(4/3) / log(2).
Using a calculator to evaluate the right side of the equation, we find that n ≈ 1.26.
Since the number of players in a basketball team must be a whole number, we can conclude that there are approximately 2 players in the team.
In conclusion, through the use of progression formulas and calculations, we have determined that the basketball team has around 2 members. This problem illustrates the concept of progressions in mathematics and how they can be applied to real-life situations, providing an engaging and challenging problem for students to solve. 進展是數學中一個迷人的概念，在各種現實生活場景中經常出現。一個例子是小明在他學校籃球隊中面臨的進展問題。
在他的第一次練習中，小明發現整個隊伍都在玩一個叫做“巧克力進展競跑”的遊戲。這個遊戲的規則很簡單：每個隊員都開始有一塊巧克力。每輪結束後，每個人都要吃掉一半的巧克力，並將剩下的一半交給隊友。小明注意到，他最初的巧克力重8克，吃掉一半後給隊友的巧克力比他手裡的重2克輕。
為了確定籃球隊中有多少個球員，我們可以使用遞迴公式來建立一個數學模型。假設團隊中有n個球員。
在第一輪中，小明收到一塊重8克的巧克力。吃掉一半後，他把一塊重(8/2 – 2)克的巧克力給了隊友。在第二輪中，巧克力的重量將是((8/2 – 2)/2 – 2)克，以此類推，直到第n輪，巧克力的重量為0克。
為了找到解答，我們需要找到使巧克力重量等於0克的n的值。通過設置這個方程，我們可以解出n：
((8/2 – 2)/2 – 2)/2 – 2)…/2 – 2 = 0。
簡化這個方程，我們有：
(8/2^n – 2^n – 2)/2 – 2 = 0。
將兩邊乘以2，我們得到：
8/2^n – 2^n – 2 – 4 = 0。
重新排列這些項，我們有：
8 – 2^(n+1) – 2^n – 4 = 0。
合併同類項，我們得到：
8 – 3 * 2^n – 4 = 0。
兩邊減去4，我們有：
4 – 3 * 2^n = 0。
兩邊除以3，我們得到：
4/3 – 2^n = 0。
解出n，我們找到：
2^n = 4/3。
取了雙方的對數，我們有：
n * log(2) = log(4/3)。
兩邊除以log(2)，我們得到：
n = log(4/3) / log(2)。
使用計算機計算方程的右邊，我們找到n≈1.26。
由於籃球隊的人數必須是整數，我們可以得出結論，這個隊伍大約有2個球員。
總之，通過使用進展公式和計算，我們確定了這個籃球隊大約有2名成員。這個問題說明了數學中進展的概念以及它們如何應用於現實生活中，為學生提供了一個有趣而具有挑戰性的問題來解決。
補中文
中文補習
補英文
英文補習
補數學
數學補習
補Econ
Econ補習
補bafs
bafs補習
暑期班
試堂優惠
豎琴課程
豎琴班
學豎琴
補ICT
ICT補習
補習ICT
補中史
中史補習
補習中史
補世史
世史補習
補習西史
補englit
補英國文學
補習Englit