1)In a basketball team,a goal made beyond the three-points,while a goal made within the three-point line counts as 2 points. If happy basketball team makes 30 goals in a game and their total score is less than 75 points, how many 3-point goals have they made at most？
我思考方向是不等式，左邊的沒頭緒。__>75 是嗎？
x+y=30 ------- (1)
2x+3y>75 -------(2)
猜的，但不懂計。
2)in the figure,L1 is the straight line with slope -1/2 passing
through a(6,2)and b.
2a)find coordinates of b
2b)if c is the mid-point of the origin and b.find the coordinates
of c
2c)Harry claims that ac is perpendicular to L1, do you agree ? explain your answer.
用了公式+設方程, 但得出未知數。而且計不到。
3)in the figure,the true bearing of city c form a and b are 070° and 160° respectively. A train travels at a speed of 120km/h from city a to cite c,while another train travels at a speed of 160km/h from city b to.cite c.After 2 hours , the two trains arrive city c.
3a)find ac and bc
3b)find angle Acb
3c)hence,find the bearing of b form a( give your answer correct to the nearest degree.)
speed*2但answer不乎
想用畢氏定理，但我answer有問題
數學功課(思考過但想不到下一步)請指教(3題)thx
2014-12-29 12:42 pm
回答 (2)
2014-12-30 12:56 am
✔ 最佳答案
(1)Let x be (2-points), y be (3-points)
x+y=30...(1)
2x+3y<75...(2) (Less than 75)
From(1),
x=30-y...(3)
Sub (3) into (2)
2(30-y)+3y<75
60-2y+3y<75
y<15
So, they made 14 3-point goals at most
(2)
AS THER IS NO INFORMATION ABOUT WHERE b IS LYING ON AND ITS RELATIONSHIP WITH L1 AND a. SO, THERE IS NO SOLUTION FOR b.
MAYBE YOU MISS SOME INFORMATION.
I JUST SHOW IF b IS LYING ON THE x-axis AND y-axis.
In the following questions, I use b(x,y)
(a)
Let b(x,y)
Construct two points P and Q which lie on the x-axis and y-axis.
Let P(x(P),0) and Q(0,y(Q))
(y-2)/(x-6)=(-1/2)
2y-4=-x+6
x+2y-10=0...(1)
Sub Q into (1)
0+2(y(Q))-10=0
y(Q)=5
So, Q(0,5)
Sub Pinto (1)
x(P)+2(0)-10=0
x(P)=10
So, Q(10,0)
(b)
Coordinates of c
=(0+x)/(2),(0+y)/(2)
(c)
First, find the slope of ac
If slope of ac*slope of L1(i.e.-1/2)=-1
Then they are perpendicular
(3)
(a)
ac=120*2=240km
bc=160*2=320km
(b)
90°
Draw the positions of the three places
Then use the reason (alt.) to find the answer
(c)
0°
2014-12-29 6:40 pm
1.
第2條式應該係2x+3y<75
x+y=30 ... 1
2x+3y<75 ... 2
因為第2條式有<而唔係=
所以只可以計算y嘅range
因此
將第1條equation 到將x做主項x=30-y
[但點解要將x=30-y放係第2條式到?
因為2x+3y<75入面,我地唔知x嘅值係幾多,所以唔可以準確計算y嘅range]
再將x=30-y放至第2條式入面
2(30-y)+3y<75
如果要將第2條式放至第1條式
咁就要轉符號
首先將x做主項x<(75-3y)/2
再放至第1條式
30=x+y<(75-3y)/2 +y
就可計算y嘅range
第2條式應該係2x+3y<75
x+y=30 ... 1
2x+3y<75 ... 2
因為第2條式有<而唔係=
所以只可以計算y嘅range
因此
將第1條equation 到將x做主項x=30-y
[但點解要將x=30-y放係第2條式到?
因為2x+3y<75入面,我地唔知x嘅值係幾多,所以唔可以準確計算y嘅range]
再將x=30-y放至第2條式入面
2(30-y)+3y<75
如果要將第2條式放至第1條式
咁就要轉符號
首先將x做主項x<(75-3y)/2
再放至第1條式
30=x+y<(75-3y)/2 +y
就可計算y嘅range
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