Xiao Ming and Xiao Li walk from village a to village B at the same time. Xiao Li's speed is 4 kilometers per hour, Xiao Ming's speed is 5 kilometers per hour. Xiao Li arrives 15 minutes later than Xiao Ming Xiao Ming and Xiao Li walk from village a to village B at the same time. Xiao Li's speed is 4 kilometers per hour, Xiao Ming's speed is 5 kilometers per hour. Xiao Li arrives 15 minutes later than Xiao Ming. Ask for the distance between village a and village B

Xiao Ming and Xiao Li walk from village a to village B at the same time. Xiao Li's speed is 4 kilometers per hour, Xiao Ming's speed is 5 kilometers per hour. Xiao Li arrives 15 minutes later than Xiao Ming Xiao Ming and Xiao Li walk from village a to village B at the same time. Xiao Li's speed is 4 kilometers per hour, Xiao Ming's speed is 5 kilometers per hour. Xiao Li arrives 15 minutes later than Xiao Ming. Ask for the distance between village a and village B

Set the distance as X km
x/4-x/5=15/60
Multiply 60 at the same time to get:
15x-12x=15
3x=15
x=5
A: the distance is 5 kilometers

Xiaoli's home is 810 kilometers away from school. She has to walk 14 hours to school every day. How many kilometers does Xiaoli walk per hour?

810 △ 14 = 3.2 (km / h); a: Xiaoli walks 3.2 km per hour on average

In the quadrilateral ABCD, the ratio of the degrees of the external angles of ∠ a ∠ B ∠ C ∠ D is 4:7:5:8, and the degrees of the internal angles are calculated

The sum of the internal angles of a quadrilateral is 360 degrees, so a = 360 * 4 / (4 + 7 + 5 + 8) = 60
B=360*7/(4+7+5+8)=105
C=360*5/(4+7+5+8)=75
D=360*8/(4+7+5+8)=120

When a voltmeter is used to directly measure the open circuit voltage of an active two terminal network, should the internal resistance of the voltmeter be larger or smaller?

Of course, the greater the internal resistance, the better. Because the open circuit voltage measurement requires that the addition of the measuring instrument does not affect the original state of the signal, if the internal resistance of the voltmeter is not large enough, it will cause the open circuit voltage to change, resulting in incorrect measurement results, so try to choose the voltmeter with infinite internal resistance for measurement

The solution of one variable linear equation in teaching and learning of grade one mathematics in junior high school (3)

Linear equation with letter coefficients
Teaching objectives
1. Make students understand and master the equation of one variable and its solution with letter coefficient;
2. Understand the meaning of formula deformation and master the method of formula deformation;
3. Improve students' ability of calculation and reasoning
Key and difficult points of Education
Key points: linear equation with letter coefficient and its solution
Difficulties: the use of letter coefficient conditions and formula deformation
Teaching process design
1、 Introduction of new courses
Q: what is equation? What is one variable linear equation?
A: an equation with an unknown number is called an equation. An equation with an unknown number and the degree of the unknown number is 1 is called a linear equation with one variable
Example solution equation 2x-13-10x + 16 = 2x + 14-1
If you remove the denominator and multiply both sides of the equation by 12, you get
4(2x－1)－2(10x+1)=3(2x+1)－12,
To get rid of the brackets, you have to
8x－4－20x－2=6x+3-12
Transfer, get
8x－20x－6x=3-12+4+2,
Merge the similar items to get
－18x=－3,
Divide both sides of the equation by - 18, and you get
X = 3.18, i.e. x = 1.6
2、 New lesson
1. The solution of linear equation with letter coefficient
We express the equation of first degree with one variable in general form as
ax=b (a≠0),
Where x is the unknown number, a and B are the known numbers represented by letters. For the unknown number x, the letter A is the coefficient of X, which is called letter coefficient, and the letter B is a constant term
If the coefficients in a linear equation of one variable are expressed by letters, then the equation is called a linear equation of one variable with letter coefficients
Second order equation
If there is no special explanation in the future, in the equation with letter coefficients, a, B, C are generally used to represent the known number, and X, y, Z are used to represent the unknown number
The solution of linear equation with letter coefficient is the same as that of linear equation with number coefficient
It should be noted here that the value of this formula can not be equal to zero when the two sides of the equation are multiplied or divided by a formula containing letters. For example, (m-2) x = 3, only when m-2 ≠ 0, that is, when m ≠ 2, can x = 3 m-2 be obtained. This is an important difference between the equation containing letter coefficients and the equation containing only number coefficients
Example 1 solves the equation AX + B2 = BX + A2 (a ≠ b)
Analysis: the letters A and B in this equation are known numbers, and X is an unknown number. It is a one variable linear equation with letter coefficients. The condition a ≠ B given here is the key to make the equation have a solution. This condition should be used in the process of solving the equation
By solving the shift term, we get
ax－bx=a2－b2,
Merge the similar items to get
(a－b)x=a2－b2.
Because a ≠ B, so A-B ≠ 0. Divide both sides of the equation by A-B, we get
x=a2－b2 a－b=(a+b)(a－b) a－b,
So x = a + B
It is pointed out that:
(1) A ≠ B is given in the problem. In the process of solving the equation, it is ensured that the solution of the equation obtained by removing the two sides of the equation with A-B which is not equal to zero is the solution of the original equation;
(2) If the solution of the equation is fractional form, it should be reduced to the simplest fraction or integral form
In case 2, x-ba = 2-x-ab (a + B ≠ 0)
Observe the characteristics of the equation structure, please say the solution of the equation
A: this equation contains fractions. You can go to the denominator first and transform the equation into a linear equation with letter coefficients
In the deformation of the equation, the known condition a + B ≠ 0
The denominator of the solution is removed and both sides of the equation are multiplied by ab
b(x－b)=2ab－a(x－a),
To get rid of the brackets, you have to
bx－b2=2ab－ax+a2,
Transfer, get
ax+bx=a2+2ab+b2
Merge the similar items to get
(a+b)x=(a+b)2.
Because a + B ≠ 0, x = a + B
It is pointed out that ab ≠ 0 is an implicit condition because the letters A and B are the denominators of two fractions in the equation, so a ≠ 0, B ≠ 0, ab ≠ 0
Example 3 solving the equation about X
a2+(x－1)ax+3a=6x+2(a≠2,a≠－3).
The solution transforms the equation into a solution
a2x－a2+ax+3a=6x+2,
Transfer items, merge similar items, get
a2x+ax－6x=a2－3a+2,
(a2+a－6)x=a2－3a+2,
(a+3)(a－2)x=(a－1)(a－2).
Because a ≠ 2, a = - 3, so a + 3 ≠ 0, A-2 ≠ 0
x=a－1 a+3.
2. Formula distortion
In physics class, we learned a lot of physical formulas. If q is the combustion value and M is the mass of fuel, then the heat w generated by complete combustion of these fuels is w = QM. For another example, the relationship between Q is the amount of electricity passing through a variant cross section, t is the time, I is the current passing through the conductor, and I = QT, If I and Q are used to express T, that is, if I and Q are known, then t = Qi
As above, transforming a formula from one form to another is called formula deformation
Taking a certain letter in the formula as an unknown quantity and other letters as a known quantity, finding the unknown quantity is to solve the problem of containing letters
In other words, formula deformation is actually solving equations with letter coefficients. Formula deformation is very important not only in mathematics, but also in physics and chemistry. We should master the skill of formula deformation
Example 4 in the formula v = V O + at, V, V O, a is known, and a ≠ 0, t is obtained
Analysis: given V, V O and a, find t, that is to say, take V, V O and a as known quantities and solve the equation of letter coefficient of unknown quantity t
By solving the shift term, we get
υ－υ0=at.
Because a ≠ 0, if you divide both sides of the equation by a, you get
t=υ－υo a.
Example 5 in the trapezoidal area formula s = 12 (a + b) h, it is known that a, B and H are positive numbers
(1) S, a, B denote h; (2) s, B, H denote a
Q: (1) which of (2) are known quantities and which are unknown quantities;
Answer: (1) s, a, B are known quantities, h is unknown quantity; (2) s, B, h are known quantities, a is unknown quantity
The solution (1) is to multiply both sides of the equation by 2
2s=(a+b)h.
Because a and B are both positive numbers, a ≠ 0, B ≠ 0, that is, a + B ≠ 0. Divide both sides of the equation by a + B, and the result is
h=2sa+b.
(2) If you multiply both sides of the equation by two, you get
2s=(a+b)h,
It's time to tidy up
ah=2s－bh.
Because h is a positive number, H ≠ 0. Divide both sides of the equation by H, and we get
a=2s－bh h.
It is pointed out that the problem is to solve the equation about h, (a + b) can be regarded as the coefficient of unknown quantity h, and (a + b) H should not be expanded in the operation
3、 Classroom exercises
1. Solve the following equation about X
(1)3a+4x=7x－5b; (2)xa－b=xb－a(a≠b)；
(3)m2(x－n)=n2(x－m)(m2≠n2);
(4)ab+xa=xb－ba(a≠b);
(5)a2x+2=a(x+2)(a≠0,a≠1).
2. Fill in the blanks
(1) If y = RX + B R ≠ 0, then x=_______ ;
(2) If f = ma, a ≠ 0, then M=_________ ;
(3) Given ax + by = C, a ≠ 0, then x=_______ .
3. The letters in the following formula are not equal to zero
(1) Find out n in the Formula M = PN + 2;
(2) Given XA + 1b = 1m, find X;
(3) In the formula s = a + B2H, find a;
(4) In the formula s = ν ot + 12t2x, find X
answer:
1.(1)x=3a+5b 3； (2)x=ab; (3)x=mn m+n; (4)x=a2+b2 a－b (5)x=2a.
2.(1)x=y－b r； (2)m=Fa; (3)x=c－by a.
3.(1)n=p－2m m; (2)x=ab－am bm; (3)a=2s－bh h;
(4)x=2s－2υott2.
4、 Summary
1. The solution of the equation with letter coefficients is the same as that of the equation with only number coefficients, but we should pay special attention to that when we multiply or divide the two sides of the equation by the equation with letter coefficients, the value of the equation can not be zero. Our examples and the conditions given in the exercises in class guarantee this
2. For the deformation of the formula, we must first make clear which is the known quantity and which is the unknown quantity
The process of finding unknown quantity is the process of solving the equation about letter coefficient
5、 Homework
1. Solve the following equation about X
(1)(m2+n2)x=m2－n2+2mnx(m－n≠0)；
(2)(x－a)2－(x－b)2=2a2－2b2 (a－b≠0);
(3)x+xm=m(m≠－1);
(4)xb+b=xa+a(a≠b);
(5)m+nx m+n=a+bx a+b(mb≠na).
2. In the Formula M = D-D 2L, all letters are not equal to zero
(1) If m, l, D are known, then D can be obtained; (2) if m, l, D are known, then D can be obtained
3. In the formula s = 12n [A1 + (n-1) D], all the letters are positive numbers, and N is an integer greater than 1
answer:
1.(1)x=m+n m－n; (2)x=－a+b 2; (3)x=m2 m+1; (4)x=ab; (5)x=1.
2.(1)D=2lM+d; (2)d=D－2lM.
3.d=2S－na1 n(n－1).
Explanation of classroom mathematics design
1. It is difficult for students to accept the understanding and solution of the equation with letter coefficient and the deformation of the formula
The relationship between the equation with mother coefficient and the equation with only number coefficient is general and special
When the letter in gives a specific number, it is an equation with only number coefficient. So in the teaching design, it is from the review to solve the equation with only number coefficient
This paper starts with the first-order equation of one variable with digital coefficients, and then discusses the solution and formula deformation of the first-order equation of one variable with letter coefficients,
It embodies the students' thinking mode from concrete to abstract and from special to general
2. In algebra teaching, we should pay attention to the infiltration of reasoning factors. In the process of solving the linear equation with letter coefficients and formula deformation, we should guide students to pay attention to the known conditions in the given problems, and correctly use the known conditions in the equation deformation. For example, in solving the equation, we often use the formula with letter to multiply (or divide) the two sides of the equation, and discuss how to use the known conditions according to the given conditions, To ensure that the value of this formula is not equal to zero, we should consciously train and improve students' logical reasoning ability, and combine algebraic operation with reasoning

How long can it take to drive 1000 watts per hour with four groups of 24 volt and 60 amp batteries to change 220 volt,

24VX60Ah=1.440KWh
1440KWhX4=5.760KWh
760kwh / 1000W = 5.76h (hour)
If inverter efficiency = 80%
Then 5.76x0.8 = 4.6 hours
According to different conversion efficiency, battery quality and electrical appliances, the actual use time will be different

As shown in the figure, the parabola y = x2 + BX + C intersects with the x-axis at two points a (- 1,0) and B (3,0), and the straight line L intersects with the parabola at two points a and C, where the abscissa of point C is 2. (1) find the analytical formula of the parabola and the analytical formula of the straight line AC; (2) P is a moving point on the line AC, and make a vertical line of the x-axis through point P