How to solve the equation with 0.36x + 4.2x = 11.4?

How to solve the equation with 0.36x + 4.2x = 11.4?

0.36x+4.2x=11.4 4.56x=11.4 x=11.4/4.56=2.5
(2x + 3) * 2 = 5x-4 to solve the equation
(2X+3)*2=5X-4
4x+6=5x-4
5x-4x=6+4
x=10
(2X+3)*2=5X-4
4X+6=5X-4
4X-5X=-4-6
-X=-10
X=10
2(2x+3）=5x-4
4x+6=5x-4
4x-5x=-4-6
-x=-10
x=10
Hello, 14519223871
(2x+3)*2=5x-4
4x+6=5x-4
5x-4x=6+4
x=10
Let a be the integer part of the root 24-1 and B the decimal part of the root 24, and find the value of a-b
The root 24-1 is about three o'clock, so a = 3
The integer part of root 24 is 4, so the decimal part is root 24-4
So A-B = 3 - (radical 24-4) = 7-radical 24
A = [(√ 24) - 1] integral part = 3,
B = [√ 24] decimal part = √ 24-4 = 2 √ 6-4.
a-b=3-(2√6-4)=7-2√6。
{x=y-3 {x+y=11 {2x+y=13 {3x-2y=8
1.{x=y-3 {x+y=11
2.{2x+y=13 {3x-2y=8
(1){x=y-3 {x+y=11
y-3+y=11
2y=14
Y=7
x=y-3=7-3=4
So the solution of the original equations is x = 4, y = 7
(2){2x+y=13 {3x-2y=8
y=13-2x
3x-2(13-2x)=8
3x-26+4x=8
7x=34
x=34/7
y=13-2x=23/7
If the image of the function f (x) = cos (3x + φ) is symmetric about the origin, then the equation of the axis of symmetry of F (x) is
If f (x) is symmetric about the origin, then f (x) is an odd function, then there is: φ = k π + π / 2, f (x) = cos (3x + K π + π / 2). When k is odd, f (x) = sin3x, K is even, and f (x) = - sin3x, so f (x) = ± sin3x, so the axis of symmetry is 3x = k π + π / 2, that is, x = k π / 3 + π / 6
On origin symmetry, then f (0) = 0
f(x)=cosφ
φ=π/2
So f (x) = cos (3x + π / 2) = - sin3x
The axis of symmetry is - sin3x = ± 1
So 3x = k π + π / 2
That is, x = k π / 3 + π / 6
Please take it_ ∩)O
Please take it~
How to convert a root to a decimal? For example, the root 8125, please give the specific process,
It's usually the square of the root fraction
With the carbon calculator, I mean the brain. Ask: what are the specific methods?
2X + y = 3,3x + 2Y = 1 [find x.y]
2x+y=3①
3x+2y=1②
①×2-②
4x-3x=6-1
X=5
∴y=3-2×5=-7
That is: the solution of the equations is x = 5; y = - 7
The minimum value of quadratic function y = x2 + X-5 is, and the value of independent variable x is
Quadratic function = 2 (x - 32) 2 + 1 the symmetry axis of the image is
2. The minimum value of quadratic function y = x2 + X-5 is, and the value of independent variable x is
3. If the function is quadratic, then the value of M is
4. If the vertex (1, - 1) of the quadratic function image is known and the image passes through the point (0, - 3), then the analytic expression of the quadratic function is
5. It is known that the quadratic function y = (m-1) x2 + 7 of X, when y decreases with the increase of X, then the value range of M is
6. If it is a quadratic function of X, then a =
7. It is known that if the image of quadratic function of X passes through the origin, then M =. Then y increases with the increase of X
8. If the image of quadratic function y = AX2 + BX + C + (A0) is shown in the figure, then point P (2a-3, B + 2)
In the coordinate system in the fourth quadrant
9. The quadratic function y = (x-1) 2 + (x-3) 2 reaches the minimum value when x =
The vertex of the parabola is C (2,), it intersects with X axis at two points a and B, and their abscissa are two of the equation, then =
2.1 recognition of function
1. Definition of function: generally, in a changing process, if there are two variables X and y that have no definite value for the independent variable x in a certain range, and y has a unique value corresponding to it, then y is said to be a function of X
2. The representation of functions: analytic method, list method and image method
3. There are several situations to determine the value range of independent variable
(1) The expression is integral, generally takes all real numbers; (2) if there is a fraction in the expression, the denominator of the fraction is not zero; (3) if there is a quadratic radical in the expression, the value of the opened way is non negative; (4) if there are multiple expressions, the common part of the value that makes each expression meaningful should be taken
2.2 quadratic function
1. The expression of quadratic function y = ax & sup2; + BX + C (a ≠ 0A, B, C are constants)
2. To understand the concept of quadratic function, we should pay attention to the following points
(1) Quadratic function is expressed by quadratic polynomials with independent variables, so the value range of independent variables is all real numbers, but when the independent variables represent the actual meaning, they are not necessarily all real numbers. (2) functions like y = ax & sup2; + BX + C are not necessarily quadratic functions, they can only be called quadratic functions if a ≠ 0
2.3 quadratic function
The images and properties of y = ax & sup2;; y = ax & sup2; + K; y = a (X-H) & sup2;; y = a (X-H) & sup2; + K; y = ax & sup2; + BX + C are as follows
image
Axis of symmetry
Vertex coordinates
nature
y=ax²
parabola
Y axis (line x = 0)
（0,0）
When a ＞ 0, the opening of the parabola is upward, and the vertex is its lowest point. On the left side of the symmetry axis, y decreases with the increase of X, and on the right side of the symmetry axis, y increases with the increase of X. when x takes the abscissa of the vertex, the ordinate of the vertex is the minimum value of Y. when a ＜ 0, the opening of the parabola is downward, and the vertex is its highest point, When x takes the abscissa of vertex, the ordinate of vertex is the maximum value of Y
y=ax² +k
parabola
Y axis (line x = 0)
(0,k)
y=a(x-h)²
parabola
Straight line x = h
(h,0)
y=a(x-h)²+k
parabola
Straight line x = h
(h,k)
y=ax² +bx+c
parabola
Line x=
1. If they are all of the same shape (Supa X; + 2) & supy; + k = Supa X; supy; + 2) & supy; + 2
2. If the image of function y = ax & sup2; is translated up or down, the image of function y = ax & sup2; + k is obtained; if the image of function y = ax & sup2; is translated left or right, the image of function y = a (X-H) & sup2; is obtained; if the image of function y = ax & sup2; is translated up or down, and then left or right, the image of function y = a (X-H) & sup2; + k is obtained
3. Parabola is composed of numerous points. To see the moving direction of the whole parabola, we can generally start from the moving direction of some special points, so as to get how the whole graph moves
2.6 determine the expression of quadratic function
1. The steps to determine the expression of quadratic function are as follows:
(1) Let the analytic expression of quadratic function (2) substitute the known points into (3) solve the equation or equations to obtain the values of a, B and C (4) substitute the values of a, B and C into the analytic expression of the function
In general, there are the following methods to find the function expression according to some conditions
(1) If three points on the parabola are known, let the analytic formula of the parabola be y = ax & sup2; + BX + C
(2) If the vertex coordinates of the parabola are known, let the analytic formula of the parabola be y = a (X-H) & sup2; + K
(3) If the coordinates of the two intersections of the parabola and the x-axis are known to be (x1,0) (x2,0), let the analytical formula of the parabola be y = a (x-x1) (x-x2)
2.7 quadratic function and quadratic equation of one variable
1. When the image of the quadratic function y = ax & sup2; + BX + C has an intersection with the X axis, the coordinate of the intersection is the value of the independent variable x when y = 0, that is, the root of the quadratic equation AX & sup2; + BX + C = 0
2. The intersection of parabola y = ax & sup2; + BX + C and X-axis:
(1) When △ = B & sup2; - 4ac > 0, the equation AX & sup2; + BX + C = 0 (a ≠ 0) has two unequal real roots, and the parabola and X axis have two intersections
(2) When △ = B & sup2; - 4ac = 0, the equation AX & sup2; + BX + C = 0 (a ≠ 0) has two equal real roots, and there is only one intersection point between the parabola and the x-axis, that is, the vertex is on the x-axis
(3) When △ = B & sup2; - 4ac ＜ 0, the equation AX & sup2; + BX + C = 0 (a ≠ 0) has no real root, and the parabola has no intersection with the x-axis
3. If the coordinates of the intersection of the parabola y = ax & sup2; + BX + C and X axis are a (x1,0), B (x2,0), then the abscissa of the intersection of the parabola and X axis is the two roots of the quadratic equation AX & sup2; + BX + C = 0 (a ≠ 0), and vice versa
2.8 application of quadratic function
1. The application of quadratic function to solve practical problems should pay attention to the correct analysis and grasp of the equivalence relationship in practical problems, and then use the equivalence relationship to list the function relationship, and finally use the function knowledge to solve problems according to the requirements of the topic
2. The basic idea of using quadratic function to solve the problem of geometric figure is: (1) according to the figure, analyze the constant and variable in the problem and the relationship between them; (2) list the geometric relationship related to the constant and variable by using geometric knowledge; (3) replace the constant and variable into the relationship, Then the equation is transformed into a function relation; (4) using function knowledge to solve problems related to geometry
If the decimal part of root 2 is a, the decimal part of root 3 is B
Finding the value of AB + A + B
a=√2-1,b=√3-1
ab+a+b=（√2-1）（√3-1）+√2-1+√3-1=√6-1
Because 4
A = (radical 2) - 1, B = (radical 3) - 1
About 1.449489742783178
If y = 2x-3, y = 2x-1
x= -5 y=-9