Math problems, please help, kneel and beg masters

Class six (1) has two-thirds more girls than boys, and the ratio of boys to girls is (3 2), and class six (2) has one-fifth more boys than girls, and the ratio of boys to girls is (6 5).

Put together three cubes with an edge length of 1 decimeter into a cuboid, the volume of this cuboid is (3) cubic decimeters, and the surface area is (16) square decimeters.

The sum of the edges of a cuboid is 48 decimeters, and the ratio of length, width and height of the cuboid is 3:2:1, then the surface area of the cuboid is (88) square decimeters.

The volume is (3) cubic decimeters, and the surface area is (14) square decimeters.

The surface area of this cuboid is (88) square decimeters.

The ratio of the radius of the two circles is 3:2, the ratio of their circumference is (3:2), and the ratio of area is (9:4).

2.3. The surface area is 1x2+1x3x4=14348x3 6=24, 48x2 6=14, 48x1 6=7, so 24, width 14, height 7

So the area is (24x14+24x7+7x14)x2=12044,6:4, area:9:4

Three-two, six-fifth.

From o do op parallel ad cross cd to p

cd=ab=8

op adcad= cop, cda= cposo, cop cad

op/ad=oc/ac=cp/cd=1/2op=ad/2=3，cp=cd/2=4

op bc, poe= cfe, ope= fceso, ope fce

CE PE = CF OP = 8 3, CE = 8CP 11 because CE + PE = CP=

1) x 10 when 20 x 30.

Solution: Let y=kx+b [take (0,,8)] Solution: Let y=kx+b [take (30,,8)].

b= {30k+b=

10k+b=8 20k+b=8

The solution gives k=, b= and the solution gets k=, b=

y= ∴y=

2)y= y=

When y=. x=6 x=25

25-6 = 9 (points).

A: Lasts 9 minutes.

If the number of lamps is t, and the price is at least set at x yuan, then x yuan will be solved.

Therefore, the price is at least set at the yuan so as not to lose money. (The smallest unit of RMB is cents).

I feel that this topic lacks conditions, and if this is the case, it should be meta).

Speaking of broken things, who wants them? Huh: D

X people in the class. Three-fifths x+4 = five-sixths (x+4).

Did you make a mistake?

Pick A so you can understand that there is exactly a right triangle left after cutting.

Cut out a maximum square with a side length of 9 cm

Two right triangles remain.

The perimeter of the trapezoid is 54 cm, and the upper and lower bottoms are subtracted by 9 cm, resulting in two right-angled triangles with 9 cm of right-angled sides

The total circumference of the remaining two right triangles is still 54 cm

54-9-9-21 = 15 (hypotenuse length).

Verify that the remaining right triangle is correct: 9*9+12*12=15*15, so it is correct, so the perimeter length = 9 + 12 + 15 = 36

The square type is 9*9

So the answer is a,

This question is not very good, if you are looking for a square, you will choose A 36

If you find two triangles, there is no solution.

There are 112 in total, and if 0 is not included, the three numbers from small to large can be:

234,246,258,345,357,369,456, 468,567, 579,678,789, a total of 16 groups, each group can form 6 three-digit numbers, a total of 96.

If it contains 0, then 0 must not be in the hundreds, and the other two numbers are doubles.

There are four groups of 12, 24, 36, and 48, and the combination of 0 can form 4 three-digit numbers per group, for a total of 4*4=16.

So there are 96 + 16 = 112 in total.