28.8 = 48-2.4x (square solution) We must use a simple equation to do it, plus checking calculation How about checking?

28.8 = 48-2.4x (square solution) We must use a simple equation to do it, plus checking calculation How about checking?

28.8=48-2.4X
2.4X=48-28.8
2.4X=19.2
X=8
Checking calculation: 48-2.4 * 8 = 28.8
Because the left is equal to the right
So x = 8 is the solution of the original equation
28.8=48-2.4x
2.4x=19.2
X=8
28.8=48-2.4x
2.4x=48-28.8
2.4x=19.2
X=8
Eight
How to solve the equation like x-0.4x = 28.8?
If it is like 0.1X + 0.08x = 18, it should be expanded by 100 times. That is to say, it is a simple method to expand the small number of decimals in each term to integer solution. The problem is transformed into 10x-4x = 288, 6x = 288, x = 48
0.6x=28.8
x=48
Is that what LZ asked?
X-0.4x = 0.6x = 28.8, so x = 28.8 / 0.6 = 48
X-0.4x = 0.6x = 28.8, so x = 28.8 / 0.6 = 48
x-0.4x=28.8
0.6x=28.8
x=28.8/0.6
x=48
Known 0
3x (1-2x) = 3x-6x & sup2; = - 6 (X & sup2; - 0.5x) = - 6 (X & sup2; - 0.5x + 1 / 16-1 / 16) = - 6 (x-1 / 4) & sup2; + 3 / 8 ∵ (x-1 / 4) & sup2; ≥ 0 ∵ - 6 (x-1 / 4) & sup2; ≤ 0 ∵ - 6 (x-1 / 4) & sup2; the maximum value is 0 ∵ y = 3x (1-2x) the maximum value is 3 / 8LZ, you are right, but the sign is wrong
It is known that there are two points a and B on the curve y ^ 2 = ax, which are symmetrical about the point (1,1), and the inclination angle of the straight line AB is 45 ° to find the value of the real number a
The inclination angle of line AB is 45 degrees
Let the line be y = x + B, passing through point (1,1)
So the line is y = X
Intersection with curve a (0,0), B (x, y)
A. B is symmetric with respect to (1,1), so point B is (2,2), which is substituted into the curve to obtain a = 2
1. Let the sum of the first n terms of the positive proportional sequence {an} be Sn, we know A2 = 2, a3a4a5 = 512 (1) find A1 and Q (2) prove that the sequence {logman} (M > 0 and m ≠ 1) is the arithmetic sequence. How to find the second problem?
2. It is known that no matter what real number B takes, there is always a common point between the straight line y = KX + B and the hyperbola X & sup2; - 2Y & sup2; = 1?
3. It is known that f (x) = ax & sup2; + (B-8) x-a-ab, f (x) > 0, X belongs to f (x) when ＜ - 3 > 2
1. (1). A2 = A1 × Q, ① A3 = A2 × Q; A4 = A2 × Q × Q; A5 = A2 × Q × Q and a3a4a5 = 512, then q ^ 6 = 64. So q = 2 (q = - 2 rounding off) is brought into ① A1 = 1, so an = 2 ^ n-1 (2) BN = logm2 ^ n-1 is proved to be the definition of arithmetic sequence: bn-1 = logm2 ^ n-2 subtracting bn-bn-1 = log2, so. 2, straight line y = KX +
It is known that the line y = (a + 1) X-1 and the curve y2 = ax have exactly one common point
Simultaneous equations: y = (a + 1) x − 1Y2 = ax, eliminate y to get: ((a + 1) x-1) 2 = ax, simplify to: (a + 1) 2x2 - (3a + 2) x + 1 = 0, when a = - 1, obviously hold, when a ≠ - 1, △ = (3a + 2) 2-4 (a + 1) 2 = 0, the solution is a = 0 or − 45, so a = 0 or - 1 or − 45
Let the lengths of the three edges of a cuboid be x, y, Z respectively, and satisfy x + y + Z = 1. The total area of the cuboid is 16 / 27
(1) Finding volume function v (x)
(2) Finding the maximum value of volume V (x)
Is it necessary to use the knowledge of derivative function
Let length, width and height be y, Z and X respectively
Y + Z + x = 1, that is y + Z = 1-x
Let 1-x be greater than 0, then x is obtained
(1)x+y+z=1,
2 (XY + YZ + ZX) = 16 / 27, then XY + YZ + ZX = 8 / 27
It is reduced to X (y + Z) + YZ = 8 / 27
By taking y + Z = 1-x into the above formula, we get x (1-x) + YZ = 8 / 27, so YZ = 8 / 27-x (1-x)
So V (x) = XYZ = x [8 / 27-x (1-x)
(2) The derivative of V (x) gives V (x) '= 3x ^ 2-2x + 8 / 27
v(x)'=3(x-1/3)^2-1/27
Then let the derivative function be equal to zero, and get x = 2 / 3. Bring it into the equation to get the maximum value
x+y+z=1 x+y=1-x
The total area is
2(xy+xz+yz)=16/27
xy+xz+yz=8/27
x(y+z)+yz=8/27
yz=8/27-x+x^2
V(x)=xyz=x(x^2-x+8/27)=x^3-x^2+8/27x (0
It is known that the line y = (a + 1) X-1 and the curve y2 = ax have exactly one common point
Simultaneous equations: y = (a + 1) x − 1Y2 = ax, eliminate y to get: ((a + 1) x-1) 2 = ax, simplify to: (a + 1) 2x2 - (3a + 2) x + 1 = 0, when a = - 1, obviously hold, when a ≠ - 1, △ = (3a + 2) 2-4 (a + 1) 2 = 0, the solution is a = 0 or − 45, so a = 0 or - 1 or − 45
Given a ∈ R, the function f (x) = 1 − 1 x, X ＞ 0 (a − 1) x + 1, X ≤ 0 (1) prove that the function f (x) increases monotonically on (0, + ∞); (2) find the zero point of function f (x)
(1) Take any two real numbers x1, X2 in (0, + ∞), and X & nbsp; 1 < X2, then f (x1) − f (x2) = (1 − 1x1) − (1 − 1x2) = 1x2 − 1x & nbsp; 1 = x1 − x2x1x2. & nbsp; ∵ 0 < X1 < X2, ∵ x1-x2 < 0, x1x2 > 0. ∵ x1 − x2x1x2 < 0, that is, f (x1) - f (x2) < 0. ∵ f (x1) < f (x2). ∵ function f (x) increases monotonically on (0, + ∞). ∵ & nbsp; (2) (I) when x ＞ 0, Let f (x) = 0, i.e. 1 − 1 x = 0, and the solution is x = 1 ＞ 0. When a ＞ 1, x = 1 is a zero point of function f (x). & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (II) when x ≤ 0, Let f (x) = 0, i.e. (A-1) x + 1 = 0. (*) ① when a ＞ 1, x = 11 − a ＜ 0 is obtained from (*); X = 11 − a is a zero point of function f (x); & nbsp; & nbsp; & nbsp; & nbsp; &In conclusion, when a > 1, the zeros of function f (x) are 1 and 11 − a; when a ≤ 1, the zeros of function f (x) are 1
Curve y = x (1-ax) ^ 2, the derivative at x = 2 is 5, find the value of real number a, a = 1, find the steps, thank you
Why multiply - A in the first step? Use derivative formula. Isn't it a whole in brackets
y=x(1-ax)^2
y'=(1-ax)^2+2x(1-ax)*(-a)
=(1-ax)^2-2ax(1-ax)
=(1-ax)(1-ax-2ax)
=(1-ax)(1-3ax)
When x = 2, y '= 5
be
5=(1-2a)(1-6a)=1-6a-2a+12a^2
12a^2-8a-4=0
3a^2-2a-1=0
(3a+1)(a-1)=0
A = - 1 / 3 or a = 1