A kind of salt water contains 20% salt. Now add 50 grams of salt and 50 grams of water. Is the salt water still 20% salt? Why?

A kind of salt water contains 20% salt. Now add 50 grams of salt and 50 grams of water. Is the salt water still 20% salt? Why?

50/(50+50)*100%=50%
Add another 50 grams of salt and 50 grams of water, the salt content is not 20%, so the salt content is not 20%

When 1kg salt is put into 7kg water, the salt content of brine is ()%

When 1kg salt is put into 7kg water, the salt content of brine is (12.5)%
1÷（1+7）=12.5%

When 20g salt is put into 100g water, the salt content of brine is []

The salt content of brine is 16.7%

Solution equation: 6.25 ^ x = 9 ^ x + 15 ^ x

6 * 25 ^ x = 9 ^ x + 15 ^ X6 * 5 ^ (2x) = 3 ^ (2x) + 3 ^ x * 5 ^ X6 * 5 ^ X / 3 ^ x = 3 ^ X / 5 ^ x + 1, divide up and down by 3 ^ x * 5 ^ x, ∵ 15 ^ x ≠ 06 * (5 / 3) ^ x = (3 / 5) ^ x + 16 = (3 / 5) ^ X / (5 / 3) ^ x + (3 / 5) ^ x (3 / 5) ^ (2x) + (3 / 5) ^ X - 6 = 0, u = (3 / 5) ^ Xu & # 178; + U -

Solve equation 6 · (9 ^ x + 9 ^ - x) - 25 (3 ^ x-3 ^ - x) + 12 = 0 (x > 0)

6[(3^x)²-2+(3^-x)²]-25(3^x-3^-x)+24=0[2(3^x-3^-x)-3][3(3^x-3^-x)-8]=02*3^x-3-2*3^-x=0 3*3^x-8-3*3^-x=02(3^x)²-3*3^x-2=0 3*(3^x)²-8*3^x-3=03^x=2 3^x=3∴x=log3 2 x=1

6. 9 / 25 = 4 / 5; X is proportional to the solution of the equation

6;9/25=4/5;x
Then 6x = 9 / 25 * 4 / 5
Then 6x = 36 / 125
The solution is x = 6 / 125

64-x = 25.6

x=64-25.6
x=38.4

Solution equation: 15x-25 = x + 25
X÷0.5+X=5.1

15X-25=X+25
15X-X=285+25
14X=50
X=50÷14
X=25/7
X÷0.5+X=5.1
2X+X=5.1
3X=5.1
X=5.1÷3
X=1.7

Solving equation 12% x + 14% (40-x) = 5

12%X+14%(40-X)=5
0.12x+5.6-0.14x=5
-0.02x=-0.6
x=30

How to solve the equation

11 / 12-x = 5 / 6
x=11/12-5/6
x=1/12
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