If a number is reduced by 60%, the result is 2 / 3 of 120. Find this number

If a number is reduced by 60%, the result is 2 / 3 of 120. Find this number

120×2/3÷（1-60%）=200

A necessary and sufficient condition for the existence of solutions to a system of nonhomogeneous linear equations AX = B is_

r(A) = r(A,b)
That is, the rank of the coefficient matrix is equal to the rank of the augmented matrix

For processing a batch of machine parts, workshop a needs 24 hours to complete the processing. Workshop B first processes 11 hours, and then the two workshops cooperate for 4 hours to complete the work. How many hours does workshop B need to complete the processing alone?

It takes x hours to set up workshop B to process separately
11X11/x+4X（1/x+1/24）=1
11/x＋4/x＋1/6＝1
15/x＝5/6
If both sides of the equation multiply x, we get
5/6x＝15
x＝18
After examination, x = 18 is the solution of the original fractional equation

X/12+5=4X/3 （3X+2）/4-（5X+1）/2=2 7X+5=6（X+1）

X / 12 + 5 = 4x / 3x / 12 + 5 = 16x / 12; 15x / 12 = 5; X = 12 / 3; X = 4; 3x + 2) / 4 - (5x + 1) / 2 = 23x + 2-10x-2 = 8; 7x = - 8; X = - 8 / 7; 7x + 5 = 6 (x + 1) 7x + 5 = 6x + 6; X = 1; I'm glad to answer for you, skyhunter 002 will answer for you. If you don't understand this question, you can ask

There are 480 kg grain in warehouse A and warehouse B. if we transport 30 kg grain from warehouse A to warehouse B, the amount of grain in warehouse A is just three times that of warehouse B

A x, b y
X+Y=480
X-30=3*(Y+30)
X=480-Y
480-Y-30=3Y+90
4Y=360
Y=90
X=480-90=390
A 390kg, B 90kg

8 (5y-3.2) = 510.4

8（5Y-3.2）=510.4
40Y-25.6=510.4
40Y=536
Y=536/40=13.4

One kilogram of waste paper can produce 0.75 kilogram of paper, and 16 big trees are needed to produce one ton of paper
A total of 160 kg of waste paper were collected in the fifth grade. How much paper can be produced? How many kg? How many tons? How many big trees have been protected by the waste paper collected in the fifth grade?

160 * 0.75 = 120 (kg) = 0.12 (T)
0.12 * 16 = 1.92 (tree)

In the equal ratio sequence, Sn = 189, q = 2, an = 96, find A1 and n

an=a1*q^(n-1)
96=a1*2^(n-1)
192=a1*2^n
Sn=(a1-a1q^n)/(1-q)
189=(a1-192)/(1-2)
189=-a1+192
a1=3
192=3*2^n
64=2^n
n=6

Xiao Hong's mother is 40 years old, but she only has 10 birthdays. Why?

Leap year is set up to make up for the time difference between the annual number of days and the earth's actual revolution cycle caused by the human calendar. The year that makes up the time difference is leap year. There are 366 days in leap year, 29 days in February of leap year. Xiaohong mother's birthday happens to be February 29 of leap year, so Xiaohong mother, 40 years old, has only 10 birthdays

As shown in the figure, in the known isosceles trapezoid ABCD, the median line length of BC ‖ ad is 28cm, the perimeter is 104cm, ad is 6cm shorter than AB, then ad: ab: BC = ()
A. 8：12：15B. 2：3：5C. 8：12：20D. 9：12：19

Suppose ad = x, ∵ ad is 6cm shorter than AB, ∵ AB = x + 6, ∵ median is 28, ABCD is isosceles trapezoid, ∵ 12 (AD + BC) = 28, ∵ C (perimeter) = AD + BC + 2 (x + 6) = 28 × 2 + 2x + 12 = 104, the solution is: x = 18, ∵ ad = 18, ab = 24, BC = 2 × 28-ad = 56-18 = 38, ∵ ad: ab: BC = 9:12:19, so D