Hurry, hurry! Math problems, hurry up! Math problems

Solution: Set up X tons to be transported from A to C, and the total cost is Y yuan.

y=20x+25(200-x)+15(220-x)+22[300-(220-x)]

20x+5000-25x+3300-15x+22x+17602x+10060

According to the nature of the function, y increases with the increase of x, so when x takes the minimum value of 0 tons, y has a minimum value of 10060 yuan.

So 200 tons from A to D, 220 tons from B to C, and 80 tons from B to D.

Solution: Set up city A to transport x tons to place C.

Total shipping = 20x + 25 (200-x) + 15 (220-x) + 22 [280-(200-x)].

20x+5000-25x+3300-15x+1760+22x2x+10060

When x=0, the minimum total shipping cost is: 10,060 yuan.

A: 200 tons from A to D and 300 tons from B to C cost the least.

Let A to C transport x tons, then A to D is 200-x B to C is 220-x B to D is X+80

Shipping cost: 20 x x + 25 x (200-x) + 15 x (220-x) + 22 x (x+80).

12 * (1 + 5 9) = 56 3 tons.

A batch of cement, 12 tons were used, and the rest was 5 9 used, and this batch of cement had 56 3 tons.

Let the weight of this batch of cement be x then (x-12) 12 = 5 9 The solution gives x=56 3

The rest: 12 times 5 9

So total: 12 + 12 * 5 9 = 56 3

Set a total of x tons. After using 12 tons, (x-12) tons remain.

x-12) = 12 multiplied by five ninths.

Please refer to ** for the detailed process.

I don't know if there's anything wrong with understanding.

English math problems, I serve.

Question, I'm more anxious than you!

What is it that makes you so anxious?

School to home 2 3 km, he can walk 5 6 km?

If "it took him a third of an hour to walk five-sixths of a kilometer" is just one of his usual things, then the equation is 2 3 (5 6 1 3).

It is calculated according to "time = distance speed", and the speed is calculated with "distance time" in parentheses.

Two-thirds of the km, the problem was said at the beginning, do you still need to solve it?

The distance is two-thirds of a kilometer, are you sure the problem is correct? I guess I asked how long it would take!

Brain teasers?

The distance is constant, it's 2 3

Follow the binomial.

t(r+1)=c6(r)(-2x)^r

r=3 coefficient is c6(3)(-2) 3=-160

Question 1: Since the minimum value of 1-x(1-x) = (x-1 2) 2+3 4 is 3 4, the maximum value of f(x) is 4 3.

Question 2: From f(-2)=f(3), we can know that the axis of symmetry x=1 2, so f(x)=a(x-1 2) 2+c, and f(x) is less than or equal to 5 constantly, so we can know a<0, c=5. Finally, we can know that f(x)=a(x-1 2) 2+5, and I personally think that the conditions are not enough.

Unable to find a

Question 3: The question should be MX 2-2MX-1. According to the title, there are m<0 and b 2-4ac<0

Therefore, 4m 2+4m<0 can be solved to -10, f(3)>0, b 2-4ac>=0

Lianli solves -3 2 Question 5: Is the topic going to be changed again, you can solve it after you change it.

Question 6: Do you see if you are right?