Solve an elementary school math problem Urgent, thank you

The truck carrying coal to a nearby power station is referred to as A, and the truck carrying coal to a distant power station is referred to as B.

If A and B transport the same number of trips, then A transports less coal tonnage than B: ton) Each trip A transports less than B: ton).

B: 6 trips).

The number of trips transported by A: 4+4=8 (trips).

A and B transport coal separately: tons), tons).

should be shipped tons).

If X cars are sent to the power station near the road and X-4 cars are sent to the power station far away on a working day, there are:

x*solution x=8 sent 8* tons to a power plant.

Then A transports less coal tonnage than B: tonnage).

Each trip A transports less than B: tons).

B: 6 trips).

The number of trips transported by A: 4+4=8 (trips).

A and B transport coal separately: tons), tons).

should be shipped tons).

Let f(x,y)=lnroot(x 2+y 2)-arctan(y x) find the partial trace derivative of x for f(x,y) fx=(x + y) (x 2 + y 2).

For f(x,y) find the partial tan pose derivative of y fy = -(x - y) (x 2 + y 2).

According to the formula derived from the implicit function, we get:

y'=-fx/fy

x + y) let take (x - y).

1 2) ln(x 2+y 2) = arcy x simultaneous derivative.

2x+2y*y']/2(x^2+y^2)]=y'Bochun x-y) [x 2(1+y 2 base delay x 2)].

Hua Dan Jian: 2x 3 (1+y 2 x 2)+2yy'(x^2(1+y^2/x^2)=2(x^2+y^2)(y'x-y)

2x^3+2xy^2+2yy'x^2+2y^3y'=2x^2y'x+2y^2y'x-2x^2y-y^3

Assuming that Mr. Wang continues to type for 15 minutes, then there should be 180 words left, that is to say, Mr. Wang has typed a total of 2700-180=2520 words in 15+15=30 minutes, then Mr. Wang's average typing speed should be 2520 30=84

Suppose Mr. Wang's average number of words typed per minute is x, then the number of words that have been typed in the first 15 minutes is 15x columns to write the equation: 2700-15x=15x-180 to solve the equation to get x=96

A: Ms. Wang's average number of typings per minute is 96

I hope to adopt it in time, and if I don't understand it, ask me again

A's speed is 400 4 B's speed is 400 3 Let's start together Three minutes later, A runs exactly one lap of 400 meters B runs 400 4x3 = 300 meters.

A ran 400 meters, B ran 300 meters, the distance from B to A was 100 meters, and the distance from B to A was 300 meters.

300 (400 3-400 4) = 9 (minutes).

Question 2 is 1 in the whole book

I watched it in 4 days. Then on the first day, I looked at 1 4, and there was 1-1 4=3 4On the 2nd day, I saw (1 4-1 32) 3 4-(1 4-1 32)=1 2+1 32 On the 3rd day, I saw (1 4-1 32+1 64) and 1 2+1 32-(1 4-1 32+1 64)=1 4+3 64=9 64

So on the fourth day, I read 19/64 of the whole book

1.。。1 (1 3-1 4)=12 If it is replaced by 5 minutes and 7 minutes, it should be 1 (1 5-1 7)=6, so I do not agree with the statement that the common multiple is said.

I was just about to help you write it, when I suddenly found a major problem: the bounty turned out to be zero egg rings. So.

5 points for a correct answer and 8 points for an incorrect answer?

10*876=8760, 8760>1000, so give 100 yuan shopping coupon, that is, 8760-100=8660

They need to pay 8,660 yuan in cash.

Buy 9 units first, pay 7884 yuan, get 7 purchase coupons, and pay 876-700 = 176 yuan for the tenth unit.

A total of 8,060 yuan was paid.

Pay 8000 to get 800 shopping coupons, the total amount is 8800, 40 more than the total price;

If you pay less than 8000, then you will only get 700 shopping vouchers, even if it is 7999+700=8699<8760, spending an extra dollar will solve the problem.

Assuming that you only need to pay x yuan in cash, then x=876*10-int(x 1000*100)=8760-int(x 10), obviously x=8760 is not the optimal value, when you buy 9 units first, you can get x=8060