A very simple calculation question for everyone

Set x eggs, y duck eggs, and z goose eggs

x+y+z=100 (1)

2)2)*5-(1): 9y+24z=400

So y=(400-24z) 9

z is an integer within 0-20, substituting the above formula, y is not an integer. No solution.

Set x eggs, y duck eggs, and z goose eggs

x+y+z=100 (1)

2) Because (1) = (2), so x+y+z=

3) Simultaneous of these three equations:

1)-(3)De: (4).

2)-2*(3): (5).

Solve (4) and (5).

Set x eggs, y duck eggs, and z goose eggs

x+y+z=100 (1)

2)2)*5-(1): 9y+24z=400

So. y=(400-24z)/9=44-2z+(4-6z)/9=44-2z+（3-6z)/9+1/9

When z is greater than 0, y cannot be an integer.

So z=0 and then push it will be unsolvable.

50 eggs are 10 yuan, 15 duck eggs are 30 yuan, and 12 goose eggs are 60 yuan.

Original. If you look at the above number, you will find that the denominator of the previous fraction and the numerator of the latter fraction can be approximated.

That's the question???

Hello: 12 and 1 4-5 and 1 2

11 5 4-5 2 4

If you have other questions, click to ask me for help after this question, it is not easy to answer the question, please understand, thank you.

Good luck with your studies! Sparrow late.

Teach you an easy way :

Create a new text document, then copy all your calculations into it, and then add a = sign in front of each calculation, as shown in the figure, and then create a new excel file on the desktop, copy all the content in the previous text document to the new excel file, then the miracle happened. If you don't believe me, you can give it a try.

Screenshot of the last answer:

The teacher's homework should be done by yourself, after all, it is very simple. In the time you have been waiting for the answer, you can actually have it done.

Part=integral<0,infinity》xd(-e (-x)))=-xe (-x)|<0, infinity" + integral <0, infinity" e (-x)dx=-(0-0)-e (-x)|<0, infinity".

11 (x 2-1)=1 [(x-1)(x+1)] around x=1, 1 (x-1).

The integral divergence can be seen from the basic anomalous integral formula.

The number of yards in the picture line is the first model Zen.

This question examines the formula: (a+b)(a-b)=a-b original formula=(5+2)(5-2)(5-2) (10+ 7)(10+ 7)(10-7)(10-7)(10-7).

√5-√2）（5-2）/（10-7）（10-7）=（√5-√2）x3/3x3

CuSO4+2NaOH = Cu(OH)2 +Na2SO4 160GCuSO4 reacts with 80GnaOH to obtain Cu(Oh)2 precipitate 98g and Na2SO4 142g. Now 100 g of CuSO4 solution contains 20 g of CuSO4, so 10 g of NaOH is needed to generate Cu(OH)2 and Na2SO4. The mass of the NaOH solution is required to be 10 20% = 50 g so the solute in the final solution is Na2SO4 with a mass fraction of.