Finding the distance between parallel lines 3x + 4y-10 = 0 and 6x + 8y-7 = 0

Finding the distance between parallel lines 3x + 4y-10 = 0 and 6x + 8y-7 = 0

3x+4y-10=0
∴6x+8y-20=0
6x+8y-7=0
The distance between two parallel lines ﹤ - A and ﹤ - D
=|-7-（-20）|/10
=1.3
3x + 4y-10 = 0 is the straight line 6x + 8y-20 = 0, while 3x + 4y-10 = 0 and 6x + 8y-7 = 0 are parallel
Using the distance formula between two parallel lines
The distance between the two lines is 13 / 10
If there is a formula, d = (10-7 / 2) / root sign (3 square + 4 square) = 1.3
Find the distance between two parallel lines 3x + 4y-1 = 0, 6x + 8y + 2 = 0
It's two fifths
1. It is known that the quadratic function f (x) = AX2 + BX, AB is constant and a ≠ 0, satisfies the condition f (5-x) = f (x-3) and the equation f (x) = x has equal roots
Is there a real number m, n (M
When f (x) = 3x, the definition field and the value field are [M, n] and [3M, 3N] respectively^^
-4,0
If f (5-x) = f (x-3) and the equation f (x) = x has equal roots, 2A = - B, (B-1) square = 0 and a = - 0.5, B = 1 can be obtained
f（x）=-0.5x*x+x
Real number m, n (m)
If integers x and y satisfy the inequality x2 + Y2 + 1 ≤ 2x + 2Y, how many values of X + y?
x2+y2+1≤2x+2y
x2+y2+1-2x-2y≤0
(x-1)2+(y-1)2≤1
x+y=1,2,3
Let X and y be real numbers, and x square + XY + y square = 1, find the value range of XY
Because x square, y square must be greater than or equal to 0, we can get: xy = 0 by transforming the equation to: x square + y square = 1-xy, so: XY > = - 1. To sum up, we can get: - 1
If x vs y vs z = 2 vs 5 vs 7, then 3x + 4y-2z of 2x + 2Y + Z equals?
∵ x to y to Z = 2 to 5 to 7
Let x = 2K, y = 5K, z = 7K, K ≠ 0
Ψ 2x + 2Y + Z: 3x + 4y-2z = (3 × 2K + 4 × 5k-2 × 7K) / (2 × 2K + 2 × 5K + 7K) = (12K) / (21k) = 4 / 7
The answer is 4 / 7
If it's a question to fill in the blanks, you can take the numbers directly, just as X is 2, y is 5, Z is 7.
If the sum of quadratic coefficient and constant term in the quadratic equation AX2 + BX + C = 0 is equal to the coefficient of the first term, then one root of the equation must be ()
A. 0B. 1C. -1D. ±1
∵ in the equation AX2 + BX + C = 0, the sum of the quadratic coefficient and the constant term is equal to the coefficient of the first term, ∵ a + C = B, that is, A-B + C = 0, then one root of the equation must be - 1
9-6y-4y ∧ 2, find the value of 2Y ^ 2 + 3Y + 7, the solution is from 9-6y-4y ^ 2 = 7 to get - 6y-4y ^ 2 = 7-9, that is, 6y + 4Y ^ = 2, so 2Y ^ 2 + 3Y = 1, so 2Y ^ 2 + 3Y + 7 = 8
Given that the value of 14x + 5-21x ^ 2 is - 2, find 6x ^ 2-4x + 5
14X+5-21X^2=7(2X-3X^2)+5=-2
Launch 2x-3x ^ 2 = - 1
6X ^ 2-4x + 5 = - 2 (2x-3x ^ 2) + 5 = 7
Therefore, 6x ^ 2-4x + 5 = 7
Happy answer, I hope to help you!
Integral substitution method
14X+5-21x^2=-2，14X+5-21x^2=-7（3x^2-2x)+5=-2，3x^2-2x=1
6X^2-4x+5=2（3x^2-2x)+5=2+5=7
If real numbers x and y satisfy x − y + 1 ≤ 0, then the value range of YX − 1 is ()
A. （-1，1）B. （-∞，-1）∪（1，+∞）C. （-∞，-1）D. [1，+∞）
The feasible region is the shaded part in the graph. The geometric meaning of YX − 1 is the slope of the line connecting the point and point a (1,0) in the region. When the line passing through point a is parallel to l: X-Y + 1 = 0, the slope is k = 1; when the line passes through points a and B (0,1), the slope is k = - 1. Therefore, if there is a common point between the line passing through point a and the feasible region, there should be k > 1 or K < - 1, so YX − 1 > 1 or YX − 1 < - 1
Solving the system of linear equations 3x + Y-Z = 7 x + 2Y + 2Z = 12 x + y + Z = 25
3x+y-z=7 x+2y+2z=12 x+y+z=25
x =38,y= -60,z=47
3x+y-z=7 1）
x+2y+2z=12 2）
x+y+z=25 3）
1) + 3): 4x + 2Y = 32, get 2x + y = 16 4)
3)*2-2):x=38
Substituting 4): y = 16-2x = - 60
Substituting 3): z = 25-x-y = 25-38 + 60 = 47
The solution is x = 38, y = - 60, z = 47