Solve math problems! Answers urgently!

Rent 7 big boats.

Because the big boat is 10 5 = 2 yuan per person.

And the boat is the yuan.

That is, to rent as many large ships as possible.

If you rent a large ship all, you need 7 7*10=70 yuan, but there is a vacant seat.

If there are 6 big boats, why do you need 2 small boats, 6 * 10 + 7 * 2 = 74 yuan, 2 empty seats.

If there are 5 large boats, why do you need 3 small boats, 5 * 10 + 3 * 7 = 71 yuan, there are no empty seats, that is to say, there is no need for adjustment.

So it's the most cost-effective to rent 7 big boats.

Hope it helps

34/5=6...4 (one way = 7).

10*7=70 (yuan).

34/3=11...1 (one way = 12).

7*12=84 (yuan).

Cheapest to rent 7 large boats.

5 people, 5*7=35 (people) 35-34=1 There is still a seat where no one sits 7 means 7 boats 7*10=70 yuan A total of 70 yuan.

Solution: The unit price of a big ship is cheap, so try to take a big boat. It doesn't matter if you have 1 seat left.

7*10=70 (yuan).

A: Take 7 big boats.

Because the big boat is 2 yuan per person, and the small boat is 2 yuan per person, so try to take the big boat as much as possible.

If you take the big boat first, then: 34 5 more than 4 people, unable to take all of them; Change 5 large boats to take 25 people, and the remaining 9 people take 3 small boats, so that the cost is 50 + 21 = 71 yuan, and the cost is the least.

Upstairs in front of it makes sense, but the title doesn't say that you can't have empty seats, so it should be 7 big ships for 70 yuan, and an empty seat, but everyone has a seat.

If all the big boats are used: 7 are needed, and one empty seat is needed, totaling 70 yuan.

If all boats are used: 12 are needed, and two empty seats are needed, totaling 84 yuan.

So the more big boats you use, the better, if you don't have empty seats, you need 5 big boats, 3 small boats, and it costs 71 yuan.

So the first method is the most cost-effective.

Because it is more cost-effective to take a large boat, try to take a large boat. You can sit on seven large boats, and there is one empty seat left. So multiplying 7 times 10 equals 70. That's the least expensive thing to do.

1.When x=3 is known, the value of polynomial ax3 (3 is to the power of 3) + bx+1 is 5 to find the value of polynomial ax3 (3 to the power of 3) + bx+1 when x=-3.

Analysis: It is known that when x=3, the value of the polynomial ax3 (3 is to the power of 3) + bx+1 is 5 and there are 27a+3b+1=5 so 27a+3b=4 so -(27a+3b)=-4

So when x=-3, ax3+bx+1=-27a-3b+1=-(27a+3b)+1=-4+1=-3

2.Football games win 1 game 3 points, draw 1 point, lose a game 0 points, if team A wins 2 times the draw, draw 1 more than the loss, a total of 21 points, ask team A to win, draw, lose how many games!

Analysis: If you lose x games, you will draw x+1 games and win 2 (x+1) games.

Another win 1 game 3 points, a draw 1 point, a loss 0 points, a total of 21 points, so 3*2(x+1)+1*(x+1)=21 solution x=2

So if you lose x = 3 games, you will draw x + 1 = 4 games and win 2 (x + 1) = 8 games.

3.If x=0 is the root of the equation 2x-3n=1 about x, then n=?

Analysis: Because x=0 is the root of the equation 2x-3n=1 about x.

So 2*0-3n=1

So n=-1 3

ax3+bx+1

When x=3, the result is 5

That is, 27a + 3b = 4

Bringing -3 in gives -(27a+3b)+1=-3, so the result is -3

If the number of draws is A, then the number of losses is A-1 and the number of wins is 2A

That is, a+3*2a=21

It's a=3

1.It is known from the meaning of the question that it is an odd function, so the answer is -5

2.If you draw x games, you will win 2x games, so 3*2x+x=21 x=3

1 When x=3, a(3) 3+b*3+1=5 ==>9a+3b=4 when x=-3, a(-3) 3+b*(-3)+1=-9a-3b+1=-(9a+3b)+1

Typing into the above equation, -4+1=-3

2 If the draw is x, then the win is 2x, the loss is x-12x*3+x*1+(x-1)*0=6x+x=7x=21 ==>x=3 wins are 6 games, the draw is 3 games, and the loss field is 2 games.

3 Find x=0 into equation 2*0-3n=1 ==> n=-1 3

==> ax^3+bx=4

Since y=ax 3+bx is an odd function, when x=-3 ax 3+bx=-4

Therefore, when x = -3, the value of the polynomial ax3 (3 is to the power of 3) + bx+1 is -32The number of wins, draws and losses is x, y, z, and there are:

x=2y;y=z+1;3x+y=21

Solution: x=6, y=3, z=2

1 When we know that x=3, the value of the polynomial ax3 (3 is to the power of 3) + bx+1 is 5, so we get 27a+3b+1=5, 27a+3b=4, -27a-3b=-4, and when x=-3, the polynomial ax3 (3 is to the power of 3) + bx+1=-27a-3b+1=-4+1=3

2 Set a draw x, then win 2x, lose x-1, 3 * 2x + x = 21, x = 3, should be 6 wins, 3 draws, and at least 2 losses.

The sum of the speeds of the two cars is: 882 7 = 126 km/h.

The speed of the bus is: 126 (5+4) 5=70 kmh.

The speed of the train is: 126-70=56 km/h.

Method 1: If the speed of the passenger car is x, the speed of the truck is 4 5 x (1 + 4 5) xx7 = 882

x = 70 method two: 882 7 = 126

The speed of the passenger car is 126 (1+4 5)=70

The speed of the wagon is 70x4 5=56

Let the unit velocity be x. Then:

5x+4x）*7=882

Yields: x=14. Therefore, the speed of the passenger car is 70 km h, and the speed of the train is 56 km h.

The speed of the two cars = 882 7 = 126

Passenger car 126 * (5 9) = 70

Train 126 * (4 9) = 56

（882÷7）×4/9=56，（882÷7）×5/9=70

70 km/h for buses and 56 km/h for trucks.

Find each portion: 882 7 = 126 126 (5+4) = 14

Find the speed by the number of parts: 14*5=70 14*4=56

1. Solution: The price should increase by X yuan per kilogram.

The equation from the problem is: (10+x)(500-20x)=6000

Simplified: x 2-15x+50=0

Solution: x1=5, x2=10

From the title: shopping malls should ensure a daily profit of 6,000 yuan, and at the same time make customers get benefits. (The price increase should be smaller).

So, x=5

Answer: The price should increase by 5 yuan per kilogram.

2. The equation for x is known to have two unequal real roots.

But the equation for x has gone to **? However, the idea of solving this problem is: (1) using the discriminant formula of the root.

Because, the equation for x has two unequal real roots.

So, b 2-4ac>0 (there must be an inequality of k here, so that the range of values of k is solved).

2) make use of the relationship between roots and coefficients; If x1 and x2 are unary quadratic equations ax 2+bx+c=0 , we get 2.

then x1+x2=-b a ,x1x2=c a

11=x1^2+x2^2=(x1+x2)^2-2x1x2 =(b/a)^2-2c/a

3. Solution: There is a problem: the bottom area is: 8 2 = 4 square meters, the pool length is x meters, then, the pool width is 4 x

Because the cost of the bottom and wall of the pool is 120 yuan and 80 yuan per square meter respectively, and it is a cuboid without a cover.

So, y=2(2x+2 4 x) 80+4 120

Therefore, the functional relationship between y and x is: y=320x+1280 x+480

Question: If a+b=2, then ab or =1 under the root number push: a+b=2》2 under the root number ab

If a+b=3, then AB under the root number or = 3/2 Launch: AB under the root number of A+B》2

If a+b=6, then the root number is under ab or =3 The same can be introduced: a+b" slag digging 2 under the root number ab

Therefore, for a>0 and b>0, there is always ab under the 2 root number of a+b》

So, 320x+1280x》2 root number (320x) (1280x).

320x+1280 x》2 (16 4 64 100).

320x+1280/x》1280

So, y=320x+1280x+480 >> 1280+480=1760

i.e.: Y" 1760

So, the minimum value of y is 1760

A: The minimum cost of the pool is: 1760 yuan.

That's it!

Let y = k· x + b

Substitute x = 1 , y = 1 and x = 1 , y = 3

k + b = 1

k + b = 3

Solution: k = 2

b = 1 so y = 2· x - 1 when x = eighth, y = 0

2ap*bc=(ac-ab)(ac+ab)2ap*bc=(ac+ab)*bc

2ap-ac-ab)*bc=0

cp+bp)*bc=0

So the point p is on the bisector of BC, and the point p is the outer center of the triangle.