Everybody help, one math problem, one math problem, everybody help.

The mass of the largest melon: 10*8 (8+7+5)=4 (kg) The mass of the medium melon: 10*7 (8+7+5)=kg) The mass of the smallest melon: 10*5 (8+7+5)=kg) The money spent by Xiao Ming is yuan).

If the unripe melon is the largest, the average cost of the edible melon per kilogram is 7 (about equal, 7 6>1, loss;

If the unripe melon is medium, the average cost of the melon per kilogram that can be eaten is 7 (4 + is about equal, 14 13>1, loss;

If the unripe melon is the smallest, the average cost of the melon per kilogram that can be eaten is 7 (4 + about equals, 14 15<1, earned.

12 (11+9+4) = kilogram If the largest melon can't be eaten, then kilogram, kilogram - the yuan that can be eaten, 12 yuan, and a loss. If the smallest melon can't be eaten, then the kilogram, 12-2 = 10 kilograms, yuan, earned.

A total of 10 kilograms, their weight ratio is 8:7:5, and their weights are 4 kilograms, kilograms, kilograms.

If 4 kg of watermelon is broken, 10-4 = 6 (kg), yuan) per kilogram of watermelon is equivalent to 7 6 yuan is greater than 1 yuan, so it is a loss. If the kilogram of watermelon is broken, kilogram), yuan) each kilogram of watermelon is equivalent to 7 yuan greater than 1 yuan, so if the kilogram of watermelon is broken, kilogram), yuan) each kilogram of watermelon is equivalent to 7 yuan less than 1 yuan, so there is no loss.

Does it mean that if it is not delicious, it is not money, but it can quench your thirst.

In the case of not counting money, it is theoretically like upstairs, but in reality, no matter how you buy it, the boss makes money, and you lose money.

I picked 7 yuan, if the small watermelon is rotten, then the value of each kilogram of good watermelon I bought. Yuan, no loss;But if it is a medium watermelon and a large watermelon, the cost of each kilogram of good watermelon is greater than 1, so it is a loss.

If it's not cooked and delicious, it won't be treated as if it wasn't bought?!!What kind of logic is this?

Solution: 1, when the area of pqc is one-third of the area of the quadrilateral pabq, then the area of pqc is one-quarter of the area of abc, and because pq is parallel to ab, pqc abc, so cp:ca=1:

2. And because the hypotenuse AB of the right-angle ABC is 10cm, and the right-angle side BC is 6cm, then the right-angled side AC is 8cm, so CP=CA*1 2=4CM

The velocity of the t-point is 4 cm s, so its current motion time is 1 s

2, because of pqc abc, cp:cq=ca:cb=8:

6=4:3, and Cp=4T, so Cq=4T*3 4=3T, So, as long as Cp+Cq=Pa+Ab+Qb, the circumference of Pqc is equal to the circumference of the quadrilateral Pabq, that is, 4T+3T=(8-4T)+10+(6-3T), and the equation is solved to obtain T=24 14=12 7 seconds.

3. When PQM is an isosceles triangle, there are three cases, namely:

1) pm=qm, then as long as m is the perpendicular point of pq and ab, at this time, p can be any point on ac, because it does not contain a, c two points, so, 0<4t<8, that is, 02) pq=qm, t has a critical value, at this critical value, pq is just equal to qm, and the two are perpendicular, when t is less than the critical value, according to the principle that the shortest point to the straight line is the perpendicular line, qm will always be greater than pq, can not meet pq=qm, and t is greater than the critical value and before reaching point a, There are m points, which can satisfy pq=qm, so when t is the critical value, pq=qm, i.e.,

4t*5 4=(6-4t*3 4)*4 5, so t=24 37, that is, when 24 373)pq=pm, the principle is the same as described in 2), there is a critical value t such that pq=pm, that is, 4t*5 4=(8-4t)*3 5, t=24 37, that is, when 24 37When t<2 is <, the total presence point m satisfies pq=pm so that pqm is an isosceles triangle.

The decimal point of the answer has shifted one place to the right, enlarged by a factor of 10, and is 10 1 9 times larger than the correct answer.

The correct answer is.

Correct answer (10-1)=18

Correct answer = 18 9 = 2

Solution: The decimal point of the answer has been shifted by one place to the right.

The value is increased by a factor of 10.

Later, the answer was numerically (10-1) times more than the correct answer, and the result was 18 times greater than the original number

So correct answer=18 (10-1)=

1. There are 4 boxes of apples and 8 boxes of pears.

2. 15 games of chess and 12 games of checkers.

3. There are 15 boys and 35 girls, 4, 15 boys and 35 girls

This mainly uses the idea of multivariate equations, and I believe that if you learn this knowledge, these problems will be easy to do, hehe. Good luck with your studies.

lao zi is in junior high school and it is difficult to find answers to bad questions in elementary school.

Go to junior high school and then do these questions BA

Empty up the whole teacher.

(1) 4 boxes of apples and 8 boxes of pears.

2. 15 games of chess and 12 games of checkers.

3) 15 boys and 35 girls.

4) There are 15 boys and 35 girls.

(1) 4 boxes of apples and 8 boxes of pears.

2) 15 rounds of chess. Checkers 12 rounds.

3) 5 boys and 45 girls (this class is obviously unbalanced in terms of male and female ratios.) 4) 35 boys and 15 girls (this class is also available.)

4 boxes of apples and 8 boxes of pears.

15 games of chess and 12 games of checkers.

There are 15 boys and 35 girls.

Yo West

Turn right 100 degrees and select C

Turn 70 degrees first, right angle, 90-20 will be due east, and then turn 30 (90-60), and it's OK.

So turn 100 degrees, to the right.