x,y的取值范围构成一个11x10的点阵，其中每一行取两点构成直角边PR，共11行的取法是11xC2,10
PR位置可互换再乘以2，确定直角边PR后剩下的一个点在剩下的10行中取再乘以10
所以11xC2,10x2x10=9900。
顺便问一下奖励是个啥
p, r s, t ,u
247 An arithmethic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant. if the list of numbers shown above is an arithmethic sequence, which of the following must also be an sarithmetic sequence?
1 2p, 2r, 2s, 2t, 2u
2 p-3, r-3, s-3, t-3, u-s
3 p^2, r^2, s^2, t^2, u^2
猜答案是哪个？
248, right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallet to the x-axis. The x- and y- coordinateds of P, Q,and R are to be integers that satisfy the inequalities -4<=x<=5 and 6<= y,= 16. How many different trangles with these properties could be constructed?
我得1,100，答案为9,900。请帮忙解释一下为什么？
239 of the 200 students at College T majoring in one or more of the science, 130 are majoring in chemistry and 150 are majoring in bilolgy. if at least 30 of the students are not majoring in either c or b, then the number of students majoring in both c and b could be any number from
d. 110 to 130
我觉得这个条件：at least 30 of the students are not majoring in either c or b
是否应该改成at least 30 of the students are not majoring in both c and b才好能？
请指教！
248, right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallet to the x-axis. The x- and y- coordinateds of P, Q,and R are to be integers that satisfy the inequalities -4<=x<=5 and 6<= y,= 16. How many different trangles with these properties could be constructed?
我得1,100，答案为9,900。请帮忙解释一下为什么？
x,y的取值范围构成一个11x10的点阵，其中每一行取两点构成直角边PR，共11行的取法是11xC2,10
PR位置可互换再乘以2，确定直角边PR后剩下的一个点在剩下的10行中取再乘以10
所以11xC2,10x2x10=9900。
顺便问一下奖励是个啥
x,y的取值范围构成一个11x10的点阵，其中每一行取两点构成直角边PR，共11行的取法是11xC2,10
PR位置可互换再乘以2，确定直角边PR后剩下的一个点在剩下的10行中取再乘以10
所以11xC2,10x2x10=9900。
顺便问一下奖励是个啥
谢谢！“剩下的一个点在剩下的10行中取再乘以10”为什么不考虑顶点可能取直线的上下？为什么不考虑这个点的x轴取值范围？－4<=x<=5
先奖励一个美景，好不好？
MM，I will choose A for the first question，An arithmethic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant.
that means the five numbers in sequences should be x, x, 2x, 4x, 8x
therefore, only A fit the requirement.
谢谢！“剩下的一个点在剩下的10行中取再乘以10”为什么不考虑顶点可能取直线的上下？为什么不考虑这个点的x轴取值范围？－4<=x<=5
先奖励一个美景，好不好？
for question 2, i don't know whether i can express my logical thinking process, but I can get the result of 9900.
It does not matter whether is range if from [-4, 5] for [0,9]. therefore, we can do a little change to get the right triangle number from this area 0<=x<=9, 0<=y<=10。
let assume that P is at the most left-down point, the number should be (10+9+8...+1)*(9+8+7+...+1)=2475
because P also can be the most right-down point, the most left-up point, the most right-up point
that's the reason why we get the result 2475*4=9900
239 of the 200 students at College T majoring in one or more of the science, 130 are majoring in chemistry and 150 are majoring in bilolgy. if at least 30 of the students are not majoring in either c or b, then the number of students majoring in both c and b could be any number from
d. 110 to 130
我觉得这个条件：at least 30 of the students are not majoring in either c or b
是否应该改成at least 30 of the students are not majoring in both c and b才好能？
请指教！
B U C = B + C - BnC = 280 - BnC
the contents above told us that (B U C- BnC)>=30
therefore 280 - 2 BnC >=30
we can come out BnC <= 125
I am sorry I can get the result in a short time, maybe you can refer to the OG explanation to find more ideas. Good Luck and B.R.
MM，I will choose A for the first question，An arithmethic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant.
that means the five numbers in sequences should be x, x, 2x, 4x, 8x
therefore, only A fit the requirement.
对，我也觉得A对，可是答案给的是（1)＋（2)。还有几题我都觉得OG11的答案错了
for question 2, i don't know whether i can express my logical thinking process, but I can get the result of 9900.
It does not matter whether is range if from [-4, 5] for [0,9]. therefore, we can do a little change to get the right triangle number from this area 0<=x<=9, 0<=y<=10。
let assume that P is at the most left-down point, the number should be (10+9+8...+1)*(9+8+7+...+1)=2475
because P also can be the most right-down point, the most left-up point, the most right-up point
that's the reason why we get the result 2475*4=9900
(10+9+8...+1)*(9+8+7+...+1)=2475 不太懂，为什么要加？
x,y的取值范围构成一个11x10的点阵，其中每一行取两点构成直角边PR，共11行的取法是11xC2,10
PR位置可互换再乘以2，确定直角边PR后剩下的一个点在剩下的10行中取再乘以10
所以11xC2,10x2x10=9900。
顺便问一下奖励是个啥
明白了！
B U C = B + C - BnC = 280 - BnC
the contents above told us that (B U C- BnC)>=30
therefore 280 - 2 BnC >=30
we can come out BnC <= 125
I am sorry I can get the result in a short time, maybe you can refer to the OG explanation to find more ideas. Good Luck and B.R.
239 of the 200 students at College T majoring in one or more of the science, 130 are majoring in chemistry and 150 are majoring in bilolgy. if at least 30 of the students are not majoring in either c or b, then the number of students majoring in both c and b could be any number from
d. 110 to 130
我觉得这个条件：at least 30 of the students are not majoring in either c or b
是否应该改成at least 30 of the students are not majoring in both c and b才好能？
请指教！
先说好，这题解答完奖励就要兑现哦
这题让我想起那位朋友的老爸发明的排方格的方法。为了方便起见咱们先把所有的数字缩小10倍。现在有20个空白的方格，其中三个已经被占满（代表那30个什么课都不上的笨学生），在剩下的17个方格中，你先填满13个选化学的，然后再填入剩下的15个选生物的，你说最少有几个方格要被填两次？当然是15-(17-13)=11个，明白了吧? 同理，要想重复最多，那就先把那13个全部重填一遍，不就是13个么？
故范围是（11，13）
奖励
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