Given that three points a (3, - 2), B (1,2) C (a, 4) are collinear, the value of a can be obtained

Given that three points a (3, - 2), B (1,2) C (a, 4) are collinear, the value of a can be obtained

Vector AB = (- 2,4)
Vector AC = (a-3,6)
Because three points a (3, - 2), B (1,2) C (a, 4) are collinear
So - 2 * 6 = 4 (A-3)
-12=4a-12
The solution is a = 0

If three points a (2,2), B (a, 0), C (0,4) are collinear, then the value of a is equal to, can this line be a curve
RT

No, collinear is straight
Let AC be y = KX + B
Then 2 = 2K + B
4=0+b
So B = 4
k=-1
y=-x+4
So 0 = - A + 4
a=4

If three points a (2,2), B (a, 0) and C (0,4) are collinear, then the value of a is equal to______ ．

AB = (A-2, - 2), AC = (- 2, 2), according to the meaning of the problem, the vector AB and AC are collinear, so there is 2 (A-2) - 4 = 0, so a = 4, so the answer is 4

In the following equation containing △ and □, the △ and □, denote different natural numbers, in which we know that a number is 99, and we know that the number in is the largest
△＋□＋○＝△×□－○

Let z = 992 and y = 492, y = 99, y = 99, y = 99, y = 99, y = 99, y = 99, y = 99, y = 99, y = 99

Given that the sum of all digits of a natural number n is 1999, find the minimum value of n

In order to minimize n, we must first minimize the number of N digits,
1999 ÷ 9 = 222 …… Yu 1
So n is the minimum number:
1999…… nine
[1 followed by 222 9s]

For 99 consecutive natural numbers, the maximum is 22.5 times of the minimum, and the sum of 99 consecutive natural numbers is obtained

The smallest x, the largest x + 98
98+x=22.5x
x=4.558?

Remember 100 natural numbers & nbsp; X, x + 1, x + 2 The sum of X + 99 is a. if the sum of a is equal to 50, what is the minimum of X?

Sum a = 100x + 9900 △ 2 = 100x + 4950, if 100x + 4950 two numbers add without carry, then sum of numbers = x sum of numbers + 4 + 9 + 5 = 50, X sum of numbers = 32, X is at least 5 digits: 99950; if 100x + 4950 two numbers add t times carry, then sum of numbers = x sum of numbers + 4 + 9 + 5-9t = 50, X sum of numbers - 9t = 32, once carry, then sum of numbers = 41, minimum 199949; twice carry, then sum of numbers = 50, minimum 699899; more Carry, X digits must be more than 5. So the minimum x is 99950

-3, 5, 7, - 11, you can write an expression with a result of 24

[5+7×(-11)]÷(-3)=24

Please add, subtract, multiply and divide the four numbers 11, 12, 7 and 2 (each number is used only once) so that the result is 24. If you answer, the answer is yes_______
According to the change of the number in the formula, fill in the operation symbols and brackets: 10 ﹣ 30 ﹣ 20 ﹣ 20 ﹣ 60 ﹣ 15 = 300 ﹣ 20 ﹣ 20 ﹣ 4 = 280 ﹣ 20 ﹣ 14 ﹣ 4 = 10

12*(11-7-2)=24

Can you get 24 by mixing 2, 4, 6 and 3? Can you come up with two methods

4×6×（3-2）=24
（4+6-2）×3=24