Junior 2 Math 5 7 Inverse propositions and inverse theorems Please answer in detail, thank you 29 20

1° set 4 as the hypotenuse.

From the meaning of the question (8-x) 2+x 2=4 2

x^-16x+64=16

16x-x^2=48

x = 4 or 12

4=4, 12>4, disagree, discard;

2° Let 8-x be the hypotenuse.

From the meaning of the title. 4^2+x^2=（8-x)^2

x=38-x=5

In line with the topic. 3° Let x be the hypotenuse.

From the meaning of the title. 4^2+(8-x)^2=x^2

x=58-x=3

In line with the topic. Answer: Yes, x=3 or 5

Special statement: I am the first year of junior high school, please advise if you have a mistake!

x>44^2+(8-x)^2=x^2

x=5x<4

4^2+x^2=(8-x)^2

x=3, so x=3 or 5 is a right triangle.

There exist, and there are three cases:

1. 4 +(8-x) =x solution gives x=5, 2. 4 +x =(8-x) solution gives x=3, 3. 4 =x +(8-x), this has no solution, so when x=5 or x=3, this triangle is a right triangle.

Assuming that the right-angled side is 4 then (8-x) 2+x 2=4 2 calculates that x exists if x is positive.

Assuming that the length of the right-angled side is 8-x, then x 2+4 2=(8-x) 2 calculates that x exists if x is positive.

Assuming that the length of the right-angled side x then 4 2+(8-x) 2=x 2 calculates that x exists if x is positive.

Monday. Obviously, supermarkets 12456 work, by 5 Zhi Department Store Monday or Friday off, by 6 Zhi Department Store Monday or Thursday off, so the Department Store is closed on Monday. You know that the bank is open on Mondays.

The answer is Monday.

1 2 3 4 5 6 .

Supermarket super.

Hundred, hundred, hundred.

Bank. As can be seen from the above figure, it is obvious that the supermarket is open for one, two, four, five and six, and the department store is open for four days, and at the same time, it will not be open for three stores a day, so the bank can only be open for one day, and then according to the fifth or sixth conditions, the supermarket may only be closed for three or five days, and then the bank may only be one or five, and then you will know it by trying.

Question 7 requires you to write five trinomials, that is, one of a and a b is 5 times, not a 5 when the sum of a and b is 5

200,000 If you set x in January, then 3x+10 in February and x+10 in January

So 5x+20=120

There are x pieces of souvenirs for A and y pieces for B souvenirs.

8x+3y=950 ①

5x+6y=800 ②

Solution: x=100, y=50

2) Suppose the purchase of A souvenir x pieces, B souvenir (100-x) pieces 7500 100x + 50 (100-x) 7650 solution set of the inequality group solution is: 50 x 53

There are 4 types of restocking plans as follows:

50 pieces of A souvenirs and 50 pieces of B souvenirs.

There were 51 souvenirs of type A and 49 souvenirs of type B.

There are 52 souvenirs of type A and 48 pieces of souvenirs of type B.

There are 53 souvenirs of type A and 47 souvenirs of type B.

3) Set the profit to $y.

y=20x+30（100-x）

y=-10x+3000

k = -10 0, y decreases as x increases.

When x=50, y is the maximum, and y=2500

Answer: In the purchase plan, the maximum profit is 2500 yuan.

Solution: (1) Set up the store to buy a kind of souvenir needs a yuan, buy a kind of B souvenir needs B yuan, according to the question to get the equation:

8a+3b=950

5a+6b=800

Solve the system of equations to obtain:

a= 100 b=50

It costs 100 yuan to buy a souvenir of type A, and 50 yuan to buy a souvenir of type B;

2) If the store buys x souvenirs of type A, then there are (100-x) souvenirs of type B purchased, 100x+50 (100-x) 7500

100x+50(100-x)≤7650

Solution: 50 x 53

x is a positive integer, x=50,51,52,53 There are 4 purchase schemes, namely: scheme 1: 50 souvenirs of A are purchased in the store, and 50 souvenirs of B are purchased;

Option 2: If the store buys 51 souvenirs of type A, then 49 souvenirs of type B are purchased;

Option 3: If the store buys 52 souvenirs of type A, then 48 souvenirs of type B are purchased;

Scheme 4: If the store buys 53 souvenirs of type A, then there are 47 souvenirs of type B (3) because the profit of type B souvenirs is higher, so the more the number of types of souvenirs, the higher the total profit, and the profit is W

then w = 20x + 30 (100-x) = -10x + 3000 k = -10 0

w varies from big to small with x.

Choose to buy 50 pieces of type A and 50 pieces of type B.

Total profit = 50 20 + 50 30 = 2500 (yuan) When buying 50 pieces of A souvenirs and 50 pieces of B souvenirs, the maximum profit can be obtained, and the maximum profit is 2500 yuan.