Given that the equation of the line L1 is y = - 2x + 3, the equation of the line L2 is y = 4x-2, the line L is parallel to L1 and the intercept of the line L2 on the Y axis is the same, the equation of the line L is obtained

Given that the equation of the line L1 is y = - 2x + 3, the equation of the line L2 is y = 4x-2, the line L is parallel to L1 and the intercept of the line L2 on the Y axis is the same, the equation of the line L is obtained

Let x = 0 in L2 give intercept y = - 2. If it is parallel to L1, then slope = - 2 and equation y = - 2x-2

6.25 × 9.9 + 0.625 simple method

6.25×9.9+0.625
=6.25×9.9+6.25 ×0.1
=6.25×(9.9+0.1)
= 62.5

The analytic expression of the straight line y = KX + B intersecting with X axis and (1,0) and intersecting with y = 2x + 3 and Y axis at the same point

The intersection of y = KX + B and X axis (1,0) is y = 0 when x = 1
The K + B = 0, B = - K equation is reduced to y = kx-k
Y = 2x + 3 and Y axis intersect at point (0,3) (x = 0, y = 3)
There is - k = 3, k = - 3, so y = - 3x + 3

If a, B and C are in equal proportion sequence, then the number of intersections between the image and X axis of the function y = ax square + BX + C is?

B = AC, so b-4ac < 0, so there is no intersection

It takes two to 56 seconds for the radio wave to reach the moon and return to the earth
It takes 2 to 56 seconds for the radio wave to reach the moon and return to the earth. It is known that the radio wave propagates 3 × 10 ^ 5km per second, then the distance between the earth and the moon is

2.56/2×3×10^5=3.84×10^5kM

Given that the solution set of the quadratic inequality ax ^ 2 + BX + C > 0 is {X1 / 2}, the solution set of the inequality CX ^ - BX + a > 0 about X is obtained

We can use the Veda theorem to get: B / a = - (1 / 3 + 1 / 2), C / a = 1 / 3 * 1 / 2, then we can use a to represent B, C, and substitute it into the following equation, we can reduce a (a > 0 can be obtained from the solution set at both ends of the zero point, so the sign is not changed), we can solve the answer

If the point representing the number a on the number axis is to the left of the origin, the result of reducing 2A + | a | is?
Which of the four options A-A B-3A Ca d3a?

Because a is to the left of the origin, then a

Given the function f (x) = x ^ 3 + ax ^ 2 + B, the tangent of the image at point P (1, f (1)) is 3x + Y-3 = 0
1) Find the monotone interval of function f (x)
2) Find the maximum value of the function in the interval [0, t] (T > 0)
I know the specific steps to solve the problem, but I don't understand why we should discuss the relationship between T and 3 in the second question. 3 has nothing to do with the maximum value point. I hope the experts can reply as soon as possible

1) By topic condition
f'(1)=3+2a=-3 f(1)=1+a+b=0
The solution is a = - 3, B = 2, then f (x) = x ^ 3-3x ^ 2 + 2, f '(x) = 3x ^ 2-6x = 3x (X-2)
Then f (x) is reduced on [0,2], and increases monotonically on (- ∞, 0), (2, + ∞)
2) Because [0, t] is a space of uncertain length, when T2, the function becomes first decreasing and then increasing, then the minimum value f (2) = - 2 is obtained on x = 2, and the maximum value of the function is the larger of the two endpoints, that is, the larger of f (0) and f (T), so we need to compare the relationship between them
From F (T) > F (0), when t > 3 or T3, the maximum value is f (T) = T ^ 3-3t ^ 2 + 2,2

10A & # 178; B + 23ab & # 178; - BA & # 178; - 32B & # 178; a merge congeners

10A & # 178; B + 23ab & # 178; - BA & # 178; - 32B & # 178; a merge of the same category = 10A & # 178; b-ba & # 178; - 32B & # 178; a + 23ab & # 178; = 9A & # 178; B - 9b & # 178; a benefactor, I see that you are a unique talent in the Wulin with strange skeleton, elegant appearance and brilliant root

The range of y = log2 (x ^ 2 - + X + 2)
Speed Wait for

y = log₂(x²-x+2)
∵ x²-x+2 = (x-½)²+1¾
∴ y = log₂(x²-x+2)≥log₂(7/4)
The value range: y ∈ [Log &; (7 / 4), ∞]
y = log₂(x²+x+2)
∵ x²-x+2 = (x+½)²+1¾
∴ y = log₂(x²+x+2)≥log₂(7/4)
The value range: y ∈ [Log &; (7 / 4), ∞]