50 kg of 60% alcohol solution is prepared from 90% and 45% alcohol solution. Use X and y to represent 90% and 45% alcohol solution respectively, fill in the following table, set up equations, and calculate the values of X and Y. 90% alcohol solution & nbsp; 45% alcohol solution & nbsp; & nbsp; & nbsp; Mass of 60% alcohol solution after mixing (kg) x y 50 mass of pure alcohol (kg)______ Water content mass (kg)______

50 kg of 60% alcohol solution is prepared from 90% and 45% alcohol solution. Use X and y to represent 90% and 45% alcohol solution respectively, fill in the following table, set up equations, and calculate the values of X and Y. 90% alcohol solution & nbsp; 45% alcohol solution & nbsp; & nbsp; & nbsp; Mass of 60% alcohol solution after mixing (kg) x y 50 mass of pure alcohol (kg)______ Water content mass (kg)______

From the meaning of the question, 0.9x + 0.45y = 300, 1x + 0.55y = 20, the solution is: x = 503y = 1003, a: the mass fraction of 90% is 503kg, the mass fraction of 45% alcohol solution is 1003kg. So the answer is: 0.9x, 0.45y, 30, 0.1X, 0.55y, 20

How to divide 5 / 1 and 15 / 7

Just change 5 / 1 to 35 / 7!

A simple algorithm of multiplying 5 / 7 by 16 / 5 by 21 / 5

5/7×16×21/5
=(5/7×21/5)×16
=3×16
=48

Simple calculation of 1 / 4 × 1 / 4 + 3 / 4 × 1 / 4

1/4 × 1/4 + 3/4 × 1/4
=1/4 ×（1/4 + 3/4）
=1/4 × 1
=1/4

Five 3, five 6 and five 9 are used to form the formula equal to 10

{9-3*3+9-3*3+(6+6+6+6+6)}/{{(9/9)*9}/3}=10

0.5 + {- quarter} - {- 2.75} + Half

Change them to decimals
Become 0.5-0.25 + 2.75 + 0.5 = 3.5

5 / 17-3 / 8 + 12 / 17 =? Simple calculation

5 / 17-3 / 8 + 12 / 17
=（5/17+12/17）-3/8
=1-3/8
=5/8

Decomposition factors: one ax & # 178; - ay & # 178; two a (X-Y) - B (Y-X) + C (X-Y)

ax²－ay²
=a(x²-y²)
=a(x+y)(x-y)
a（x-y）-b（y-x）+c（x-y）
=a(x-y)+b(x-y)+c(x-y)
=(x-y)(a+b+c)

(2) 4.1 times 2.72 minus 1.53, what is the quotient of the product divided by 24.395
There should be a formula

（1）18×3/4÷1/4=54
（2）24.395÷[4.1×（2.72-1.53）]=5

With a 52 cm long lead wire, a 6 cm, 4 cm wide and () cm high rectangular teaching aid can be welded
A. 2B. 3C. 4D. 5

(52-6 × 4-4 × 4) △ 4, = (52-24-16) △ 4, = 12 △ 4, = 3 (CM); therefore, B