Here's a balance and a 30 cm scale. Can you measure the length of a roll of thin copper wire with uniform thickness

Here's a balance and a 30 cm scale. Can you measure the length of a roll of thin copper wire with uniform thickness

First, remove the 30cm copper wire and measure the mass M1 with a balance
Then weigh the total mass m of all copper wires with a balance,
Let the length be L,
According to m / L = M1 / 30, l = 30m / M1
The premise is that the balance can weigh the total mass and the measuring range is large enough

English exercise book answers

What version?
And you can't get the answer if you ask like this. You'd better send the questions. Of course, if your exercise book is famous and widely used, you can search it in Baidu Library. Generally, it's conjoined with the answers, and the answers are at the end or the center
In fact, we should do the problem by ourselves. Sixth grade English is not difficult

The least common multiple of two numbers is 120 and the greatest common factor is 8. One of them is 40 and the other is 8

120÷40×8=24
The other number is 24

List the proportions and answer: 1. The ratio of X to 10 is equal to the ratio of 30 to 20. 2. The ratio of 1 / 3 to 1 / 5 is equal to the ratio of X to 20

1.
x:10=30:20
20x=10*30
20x=300
x=15
2/
1/3:1/5=x:20
x/5=1/3*20
x=100/3

Give a counterexample to the following proposition to show that it is a false proposition
If a & # 178; > b & # 178;, then a > B

For example (- 3) &# 178; > 2 & # 178;, and - 3 < 2

Fill in the numbers according to the rules: 1,8,27,64, ()

1 = 1 to the third power, 2 = 2 to the third power... And so on & nbsp; so: & nbsp; fill in the brackets: 125, 216 & nbsp; are 5 to the third power, 6 to the third power & nbsp; ~ I will always 523 for you to answer, I wish you progress in your study ~ ~ ~ if you agree with my answer, please timely click the [adopt as a satisfactory answer] button ~ ~ mobile phone

If x + y = 10, xy = 7, then x ~ 2Y + XY ~ 2 =?

X^2Y+XY^2
=XY(X+Y)
=7*10
=70

Suitable for junior one, not too difficult, but a little difficult
It's either applied or similar to filling in the blanks

1.2(x-2)-3(4x-1)=9(1-x)
2.11x+64-2x=100-9x
3.15-(8-5x)=7x+(4-3x)
4.3(x-7)-2[9-4(2-x)]=22
5.3/2[2/3(1/4x-1)-2]-x=2
6.2(x-2)+2=x+1
7.0.4(x-0.2)+1.5=0.7x-0.38
8.30x-10(10-x)=100
9.4(x+2)=5(x-2)
10.120-4(x+5)=25
11.15x+863-65x=54
12.12.3(x-2)+1=x-(2x-1)
13.11x+64-2x=100-9x
14.14.59+x-25.31=0
15.x-48.32+78.51=80
16.820-16x=45.5×8
17.(x-6)×7=2x
18.3x+x=18
19.0.8+3.2=7.2
20.12.5-3x=6.5
21.1.2(x-0.64)=0.54
22.x+12.5=3.5x
23.8x-22.8=1.2
24.1\ 50x+10=60
25.2\ 60x-30=20
26.3\ 3^20x+50=110
27.4\ 2x=5x-3
28.5\ 90=10+x
29.6\ 90+20x=30
30.7\ 691+3x=700

A group of numbers in regular order: 1, 2, 4, 8, 16, etc., among which the number 2011 is () a, the 2011 power B of 2, the 2011 power-1 C of 2, and the 201 power of 2
Power D and above are all wrong

The zeroth power of 1 = 2
2 = 1 power of 2
4 = the second power of 2
The third power of 8 = 2
The fourth power of 16 = 2
······
The first number is the zero power of 2
The second number is to the power of 2
The third number is the second power of two
The fourth number is 2 to the power of 3
······
The nth number is the (n-1) power of 2
Therefore, the number 2011 is the (2011-1) power of 2, that is, the power 2010

Determinant problem: Solutions of X, a, B, C; a, x, C, B; B, C, x, a; C, B, a, X; = 0

c1+c2+c3+c4x+a+b+c a b cx+a+b+c x c bx+a+b+c c x ax+a+b+c b a xr2-r1,r3-r1,r4-r1x+a+b+c a b c0 x-a c-b b-c0 c-a x-b a-c0 b-a a-b x-cc2+c3x+a+b+c a+b b c0 x-a-b+c c-b b-c0 x-a-b+c x-b a-c0 0 a-b x-cr3-...