Who is good at math? In! Solve with equations or systems of equations or inequalities or inequalitie

Solution: (1) The cost paid by the customer for shopping in supermarket A is Y1, and the fee paid for shopping in supermarket B is Y2, according to the topic.

y1 = 300 +

y2 =2) makes y1 = y2, i.e. 300 + =

Solution x = 1200

Therefore, when the customer has accumulated 1,200 yuan > shopping, Home A is more discounted; When x < 1200 yuan, B is more preferential.

Solution: (1) It costs X yuan to rent a car A and Y yuan to rent a car B.

x + 2y = 2500

2x + y = 2450

Solving this system of equations gives x = 800 , y = 850

2) If you rent a model car Z, then rent a B model car (6-Z), which is derived from the title.

16z + 18(6-z) >= 100

800z + 850(6-z) <= 5000

Solution 2 <= z <= 4

When z=2, 6-z=4;When z=3, 6-z=3;When z=4, 6-z=2

Therefore, there are three types of car rental schemes, which are to rent 2 cars A, 4 cars B or 3 cars A, 3 cars B or 4 cars A, and 2 cars B.

3) Because the cost of car B is higher than that of car A, the minimum car rental fee is required, and it must be that the less car B, the lower the cost, so when car A has 4 cars and car B has 2 cars, the car rental cost is the lowest, and the minimum cost is 800 * 4 + 850 * 2 = 4900 yuan.

First question: A; (x-300)*

Second; （x-200)*

Second question: B-A = When x is greater than 600, shop A is cheaper, and when x is less than 600, shop B is cheaper.

One question: car A 800, car B 850

Second question: 800x+850y is less than or equal to 5000 and 16x+18y is greater than or equal to 100 Just solve the two inequalities.

Question 3: Just draw an interval point using two inequalities.

1: A: y=

B: y=From the knowable: when x>600 store A discount.

800x+850y<=5000

x>=0,y>=0

And from 16x+18y>=100,x+y=6,800x+850y<=5000, we can know that 2=due to w

The first question: A 300 + (x-300) * 80% = 60 + B 85% * x =

Second question: 3001200 A discount.

The first question: A and B each need x and y yuan, then x+2y=2500 and 2x+y=2450

So x=800, y=850

The second question: if A m is a vehicle, then B is 6-m a vehicle, 16m + 18 (6-m) 100 and 800m + 850 (6-m) 5000

Solution 2 m 4

Therefore, car rental plan 1: 2 cars A and 4 cars B; Car rental plan 2: 3 cars A and 3 cars B; Car rental plan 2: 4 cars A and 2 cars B.

The third question: the fare of car rental plan = 2 * 800 + 4 * 850 = 5000;Car rental plan 2: fare = 3 * 800 + 3 * 850 = 4950;Car rental plan 3 fare = 4 * 800 + 2 * 850 = 4900

In summary, the minimum car rental cost is 4900 yuan.

a+b)=a 2+b 2+ab+ba=a+b, because a 2=a b 2=b

So ab+ba=0

a 2 = a so the eigenvalue of a has.

b 2-b=0 =>b=0 or posture b=1 (b is the eigenvalue of a) ab+ba=0 left multiplication a.

ab+aba=0

ab(e+a)=0

Because the special year of a can only be selected between 0 and 1, the characteristic value of a+e can only be selected between 1 and 2.

So the a+e determinant is not equal to 0

Then a+e is irreversible, which means that there are n uncorrelated vectors.

That is to say, ab has n basic solution systems (because ab(e+a)=0, e+a can be regarded as the solution of ab's homogeneous equation).

That is, the rank of ab is 0

Then ab can only be 0

2x+1＞3，x＞1

20x-a＜6

x＜(a+6)/20

There are 3 integer solutions, which can be met by x 1 and 2, 3, 4 and because x (a+6) 20

So 4 (A+6) 20 5

80＜a+6≤100

76＜a+2≤96

Because 8x-a=2 is solved as an integer.

8x=a+2, so a+2 is a multiple of 8.

Between 76 and 96, there are 80, 88, 96 that satisfies multiples of 8, so a+2 = 80 or 88 or 96

a = 78 or 86 or 94

There are 3 A-values that meet the conditions.

College Entrance Examination Mathematics Question Analysis Question 6 Questions with a total of 70 points, this answer includes all calculations and proof questions, there must be calculation questions such as calculating trigonometric functions, proving inequalities, calculating probability expectations, etc.

1 solution: Set the original production of x vehicles per day.

then 15(x+6)>20x

x<18 Because x is an integer, the maximum integer value of x = 17 A: Originally, a maximum of 17 units were produced per day.

2.There are x rooms on the ground floor, and there are (x+5) rooms on the second floor, which is 4x<18 from the title

5x>18

4(x+5)>18

Solving inequalities gives 9 2>x>18 5 so x=4 (between) 3The actual charge of travel agency A is w=500*2+500*the actual charge of travel agency B is n=500*

When m>n is 350x+1000>400x+800x<4 (people), choose B travel agency.

When x > 4 people, choose a travel agency.

When x = 4 people, both are fine.

4.If the discount of x is calculated, then 2190x+<=2190(1+x<=, so at least 8% discount is costable.

5.(1) Let the profit after x years be 72x>120+40x x>15 4 x is an integer, so x=4

So make a profit after 4 years.

2) (72*15-120-40*15+20) 15=76 3 (10,000 yuan).

6.Set the invested funds at the beginning of October ** profit m = (1+

Profit at the end of the month n=

When m>n, i.e., when x<20000, the profit at the beginning of the month is more, and when the profit is more at the end of the month when x<20000.

Same when x=20000.

Solution: Let y be the total monthly cost, then.

In the first way, y1 = 15+

The second way, y2=

y2-y1=, when y2-y1=>0, x>150, choose the first mode;

When y2-y1=<0, 0x<150, choose the second way, when y2-y1=, x=150, both ways are acceptable.

Set the call time to be x minutes, one company pays Y1 yuan, and the second company pays Y2 yuan.

y1=15+ y2=

When y1=y2, that is, x=150, the two companies pay the same fee.

When y1 > y2, i.e. x<150, the second company is chosen.

When Y1150 is the first company.

Let the talk time be m, when the two methods are equal, it is a critical point 15+ solution: m=150

Therefore, when the call time is greater than 150 minutes, choose the form of monthly rent for less than 150 minutes, and choose the form of no monthly rent.

The first cost is 15+

The second cost is:

15+ i.e. x 150 When you choose the second payment method, 15+ i.e. x 150, when you choose the first payment method, 15+ i.e. x = 150, both payment methods are acceptable.

Let's start by considering the case where the two approaches are equal.

Find x, and then you can see according to the equation that if x increases, then "15+" is the case where the two ways are equal) If x decreases, then <15+ is the case where the two ways are equal) and then you can find the solution by case.

solution, x+2y=-5

3x-4y=25②

3x+6y=-15 is obtained

Then - gives 10y=-40

Then y=-4 brings in x=3

Set the master to win x, and the grandson to win 10-x. There are x 3 (9-x), x 3 (10-x), x is 27 4 x 15 2.

x is an integer, x=7, that is, the master wins 7, and the grandson wins 3.

3*a+8=5*a-2(1) a=5 total is 23 rounded off to get a=6 total is 26 and the same is true for others.

On a monthly basis. Pay 15+ of the monthly rent. Those who do not pay monthly rent. Because x doesn't know the exact value.

15+>。Solution x<150Therefore, if the call time is less than 150 minutes per month, it is cost-effective not to pay the monthly rent.

15+ "Solution x>150So the monthly talk time is more than 150 minutes. It is a good deal to pay monthly rent.

If the talk time is equal to 150 minutes. Both are fine.

1；15+

X<150 The latter is better.

When x=150, both are the same.

X<150 The former is better.