Xiaogang and Xiaoqiang practice running on the 400 meter circular track. Their speed is 200 meters per minute and 160 meters per minute respectively

Xiaogang and Xiaoqiang practice running on the 400 meter circular track. Their speed is 200 meters per minute and 160 meters per minute respectively

Set X minutes later, that is, Xiaogang surpasses Xiaoqiang by 1 lap
200X = 160X+400,
X = 10 minutes

Xiaoqiang height 143 Xiaogang height unknown Xiaoming height 136 average height 139 Xiaogang height

Let Xiaogang be x in height
Then the equation (143 + X + 136) / 3 = 139 is given
X=138
A: Xiaogang is 138

Xiaogang is 1.2 meters tall. Xiaoqiang is 1.1 times as tall as Xiaogang. How many meters is Xiaoqiang

1.2 * 1.1 = 1.32 (m)

Xiaoming is a centimeter tall, Xiaogang is 18 centimeters higher than Xiaoming, Xiaogang is 12 centimeters shorter than Xiaoqiang, the average height of the three is () cm?

(a + 16), wrong classification

3(5x+4)+4(x-1)=2-12(5x-5)

3(5x+4)+4(x-1)=2-12(5x-5)
15x+12+4x-4=2-60x+60
19x+8=62-60x
19x+60x=62-8
79x=54
x=54÷79
x=54/79

A cylinder is 1 meter long. After it is cut into two equal small cylinders, the surface area increases by 8 square decimeters. What is the volume of the original cylinder?
What's the volume of the original cylinder? What's the volume of this cylinder?

Cut into two small cylinders, then cut out two circular surfaces, so each circular surface, that is, the bottom area of the original cylinder is 8 / 2 = 4 square decimeters;
So the volume of the original cylinder is 10 * 4 = 40 cubic decimeter

Fill in ten numbers in 1.2.3.4.5.6.7.8.9, each number can only be used once. Ten is equal to ten, ten is equal to ten
Fill in 10 numbers 1.2.3.4.5.6.7.8.9, each number can only be used once
Mouth ten equals mouth ten equals mouth ten equals mouth ten equals mouth ten equals mouth ten equals mouth ten

1+9=10
2+8=10
3+7=10
4+6=10
5+5=10

1. The sequence {an} is an arithmetic sequence with non-zero tolerance, and A7, A10 and A15 are three consecutive terms of an arithmetic sequence {BN}. If the first term B1 of the arithmetic sequence is 3, then BN is calculated
2. The second, third and sixth items of the arithmetic sequence with tolerance not 0 constitute an arithmetic sequence in turn, and the common ratio Q of the arithmetic sequence is obtained
3. If all items are positive, the equal ratio sequence {an}, the common ratio ≠ 1, A5, a7, a8 become the equal difference sequence, and find the common ratio Q

Let A1 = a, then A7 = a + 6da10 = a + 9da15 = a + 14d, so (a + 9D) ^ 2 = (a + 6D) (a + 14d) a ^ 2 + 18ad + 81D ^ 2 = a ^ 2 + 20ad + 84d ^ 22ad + 3D ^ 2 = 0d ≠ 02A = - 3DA = - 3D / 2q = A15 / A10 = (a + 14d) / (a + 9D) = (- 3D / 2 + 14d) / - 3D / 2 + 9D) = 5 / 3bn = 3 * (5 / 3) ^ (n-1) process: let the first

It is known that the opposite number of a is 2 / 3. The reciprocal of B is negative one and one fifth. Find the value of a + 3B / a-2b

a=-2/3
b=-5/6
(a+3b)/(a-2b)
=(-2/3-5/2)/(-2/3+5/3)
=(-19/6)/(1)
=-19/6

If for any non-zero rational numbers a and B, the new definition operation #: a #, B = AB + 1, then what is the value of (- 5) # (+ 4) # (- 3)

(-5）#（+4）=（-5）×（+4）+1=-19
（-19）#（-3）=（-19）×（-3）+1=58