Who will do these two math problems, please? Help

1.(1) If the original plan is to rent x 45-seater buses, the total number of students in this batch is (45x+15).

45x+15=60(x-1) x=5 (vehicles), then 45x+15=240 (people).

2) Rent a 45-seater bus: 220 * 5 = 1100 (yuan) Rent a 60-seater bus: 300 * (5-1) = 1200 (yuan) To sum up, it is more cost-effective to rent 5 45-seater buses, a total of 1100 yuan.

2.Set the wholesale price of each pencil at X yuan, and the wholesale price of each eraser at Y yuan.

then we get the system of equations 30[2(x+

30(3x+2y)=

Then x= y=

The wholesale price of each pencil is yuan, and the wholesale price of each eraser is yuan.

Suppose a total of x quantities are required.

60x-45x=60 (one more car with 60 seats).

x=4 for a total of 240 people. 4 amounts.

Question 1: 45x+15=60x-60x=5 (vehicles), originally planned to rent 5 vehicles with 45 seats. Number of students = 45 * 5 + 15 = 240 (people).

Rent 4 cars with 45 seats for 880 yuan to carry 180 students, and rent 1 car with 60 seats for 300 yuan to carry 60 people (just all students have seats, total car rental fee = 880 + 300 = 1180 yuan).

Question 2: If the wholesale price of pencil is $x and the wholesale price of eraser is $y, then 30*2*(x+30*3x+30*2y=.)

Solve this system of equations, get x = yuan), y = yuan) That is, the wholesale price of pencils is two cents and five cents per piece, and the wholesale price of erasers is three cents per piece.

Let's say there are x cars with 45 seats.

60x-45x=60+15->x=5 cars, there are 240 students.

4 cars with 45 seats + 1 car with 60 seats, everyone has a seat.

Suppose the wholesale price of pencils is x and the wholesale price of erasers is y, then the cost per person:

3x+2y= ->3x+2y=

2(x+ -2x+ y=

x= y=

a+b)=a 2+b 2+ab+ba=a+b, because a 2=a b 2=b

So ab+ba=0

a 2 = a so the eigenvalue of a has.

b 2-b=0 =>b=0 or posture b=1 (b is the eigenvalue of a) ab+ba=0 left multiplication a.

ab+aba=0

ab(e+a)=0

Because the special year of a can only be selected between 0 and 1, the characteristic value of a+e can only be selected between 1 and 2.

So the a+e determinant is not equal to 0

Then a+e is irreversible, which means that there are n uncorrelated vectors.

That is to say, ab has n basic solution systems (because ab(e+a)=0, e+a can be regarded as the solution of ab's homogeneous equation).

That is, the rank of ab is 0

Then ab can only be 0

If the total volume of a small cube is 125, then each is 125 8, and the side length of each is 5 2, so the surface area is 150 4=

So there is x+y=1 and its cube root has a value of 1

1. The side length of each small wooden block = 125 The cubic root of 8 = 5 2 The surface area of each small wooden block x-7=-(3y+4).

x+y=1xThe cube root of y=1

Since x and y are opposite to each other.

So y = -x

Then the original becomes: {2x - x) = 3

2kx+(k+1)（-x）= 10

Simplified later get {3x = 3.)

kx - x = 10

i.e. { x = 1

kx - x = 10

Substituting x = 1 into the following equation gives k - 1 = 10 so k = 11

Solution: Because x and y are opposite numbers to each other.

So y = -x

Because 2x-y=3

So 2x-y=3

2x+x=3

x=1y=-1

Combine x=1y=-1 generation 2kx+(k+1)y=10 to get k=11

Substituting x=-y into the first equation gives y=-1, x=1, and then substituting the second equation to get k=11

The denominator is (3 2*2) 3-(3 2*4) 3+(3 2*6) 3-(3 2*8) 3+(3 2*10) 3-(3 2*12) 3=(3 2) 3*numerator.

So the result is a, 8 27

It's not easy to write, the numerator proposes the cube of common factor 2: 2 3 (1-2 3+3 3-4 3+5 3-6 3), and the denominator proposes the cube of common factor 3: 3 3 (1-2 3+3 3-4 3+5 3-6 3), divide the minute, do not count in parentheses, it is the same at first glance, approximate the minute, the remaining 2 3 3 3 = 8 27, choose a

1. This question is not large, just calculate it directly.

2. The numerator puts forward the 3rd power of 2, and the denominator puts the 3rd power of 3, and the rest is the same, so the result is 8 27