I hope that the math problem of a one dimensional quadratic equation will be completed before noon t

Set up stocking for x days.

Acquisition cost: 1000 30

Stocking fee: 400x

Number of deaths: 10x

Remaining quantity: 1000-10x

Sales revenue: (1000-10x) (30+x) + 10x * 20 profit: (1000-10x) (30 + x) + 10x * 20-1000 30-400x

Equation: (1000-10x)(30+x)+10x*20-1000 30-400x=6250

100-x)(30+x)+20x-3000-40x=625x^2+70x+3000-20x-3000=625x^2-50x+625=0

x-25)^2=0

x=25A: Sold at one time after 25 days of stocking.

x is the number of days and y is the profit.

y=(30+x)*(1000-10x)+20*10x-400x-30*1000

500x-10x^2

then 10x 2-500x+6250=0

x-25)^2=0

x = 2525 days.

1000-10x)(30+x)+200-400x=6250 Find x is the number of days of stocking.

Solution: 625*(1-20%)=500 (yuan) 500*(1+6%)=530 (yuan).

Let the cost after the drop be x

625*2-500*2=500+530-2*xx=390

Solution: Let the cost after the drop be x

625*2-500*2=500+530-2*xx=390

According to the topic, it is known that because the water consumption is 15m in two months, the water consumption in March should be about 6 tons, and the water consumption in April should be within 6 to 10 tons.

x+y=15

2x+2*6+4*(y-6)=44

Solution: x = 2 tons.

y = 13 tons.

A: 2 tons of water were used in March and 13 tons in April.

Solution: Set the water use in March XM and the water use YM in April

x+y=15①

2x+2*6+4*(10-6)+8*(y-10)=44 from , x+4y=48

, 3y=33

y=11 Substitute y=11 to get x=4

x=4, y=11 are the solutions of the original equations.

A: The water used in March is 4m, and the water is used in April is 11m.

Solution: It can be obtained by removing the parentheses on both sides of the equation at the same time.

x^2-mx-2x+2m=p^2-mp-2p+2mx^2-(m+2)x=p^2-(m+2)px^2-p^2=(m+2)(x-p)

x-p)(x+p)=(m+2)(x-p)

So x1=p x2=m+2-p

1.Set the unit price of apples to be x yuan.

30-4-2x)/3=(30+8-3x)/44(26-2x)=3(38-3x)

104-8x=114-9x

x=10 peaches unit price = (26-2*10) 3=2 yuan.

Apples and peaches are priced at 10 yuan and 2 yuan each.

It also takes x hours to set the armor, x 8 + (1 8 + 1 6) = 1

x/8+7/24=1

x/8=17/24

x=17/3

A still do 17 for 3 hours.

The first problem uses a unary equation, hehe, it's still binary simple!

2x+3y=26, 3x+4y=38, just solve it!

Question 2: 1 8*x=1-1 6, solve x out of -1!

1 Let the apple be x yuan and the peach y yuan yuan, and the columnable equation is 2x+3y+4=30

3x+4y-8=30

The solution is x=10 y=2

2. Let the total project be 1 A do 1 8 in one hour and B do 1 61-(1 8 + 1 6) * 1 = 17 24

17 24 divided by 1 8 gives 17 3 hours.

It is set to arrive in x hours.

50x+20=50(x+24/60)=75(x-24/60)=75x-30→x=(20+30)/(75-50)=2

50*2+20)/2=60。He travels at a speed of 60 kilometers per hour and arrives on time.

Set up a large car to transport x tons each time and a small car to transport y tons each time.

2x+3y= launch 4x+6y=315x+6y=35

Launch x=4 y=

3 large carts and 5 small carts can transport 12+ tons at a time).

Hope it helps.

The small car transports XT at one time, and the large cart transports YT at one time

3x+2y=

6x+5y=35

x= y=4

5x+3y=

3 large carts and 5 small carts can transport goods at one time.

Solution: Set up a large car to transport x t, a small car to transport y t, from the question to know:

2x+3y＝。。1）

5x+6y＝35。。。2）

2) Formula (1) *2 gets: x 4

Substituting equation (1) gets: y

So 3x+5y 3*4+5*