Put three rectangles that are 7 cm long and 2 cm wide together to form a big rectangle. The perimeter of the big rectangle is______ Cm or______ Cm

Put three rectangles that are 7 cm long and 2 cm wide together to form a big rectangle. The perimeter of the big rectangle is______ Cm or______ Cm

(7 × 3 + 2) × 2, = 23 × 2, = 46 (CM); (7 + 2 × 3) × 2, = 13 × 2, = 26 (CM); answer: the circumference of a large rectangle is 46 cm or 26 cm

What is the formula for calculating the area and perimeter of the bottom of a cylinder?
The formula

Area: 1 / 2 * π * circle radius perimeter: π * circle diameter

Root number 20-2 times root number 1 / 5 + root number 3 and 1 / 5

Root number 20-2 times root number 1 / 5 + root number 3 and 1 / 5
=2√5-2/5√5+√16/5
=2√5-2/5√5+4/5√5
=(2-2/5+4/5)√5
=12/5√5

Differential of y = e to the power of π - 3x * cos3x

dy=d[e^(π-3x)*cos3x]=cos3xde^(π-3x)+e^(π-3x)dcos3x=cos3x*e^(π-3x)d(π-3x)+e^(π-3x)*(-sin3x)d(3x)=cos3x*e^(π-3x)*(-3)dx+e^(π-3x)*(-sin3x)*3dx=-3e^(π-3x)(cos3x+sin3x)dx=-3√2sin(3x+π/4)*e^(π-3...

Chemical expressions
All literal expressions and symbolic expressions (PEP)

According to the element classification, the chemical equation related to oxygen is summarized as follows: 2mg + O2 ignition = = = = 2MgO phenomenon: combustion, release a lot of heat, and release dazzling white light at the same time s + O2 ignition = = = = SO2 phenomenon: light blue flame in the air; blue purple flame in pure oxygen; pungent odor generated at the same time

How can three 10's and one 7's be equal to 24
3 10 1 7

There is no answer, you can go to the following online calculation:

Given the linear function Y1 = 2x-5, y2 = - 3x + 4, if Y1 ≥ 2Y2, then the value range of X is

y1≥2y2
The results are as follows
2x-5≥2(-3x+4)
2x-5≥-6x+8
2x+6x≥5+8
8x≥13
x≥13/8

The perimeter of a circular fountain is 62.8m. How many square meters does the fountain cover?

2, = 3.14 × 102, = 314 square meters; a: it covers an area of 314 square meters

How to calculate 24 points with 3 5's and 1 1's?

Too easy (5-1 / 5) * 5 = 4.8 * 5 = 24

Finding the radius of curvature at the highest point of normal distribution curve
It's a standard normal distribution curve

Normal distribution equation
{[1/sqrt(2pi)δ]} * exp[-(x-u)^2/(2 * δ^2)]
The formula of curvature radius by derivative
R = [(1+y'^2)^(3/2)]/|y"|
The highest point of normal distribution is y '= 0
Find the second derivative and count back
We obtain r = sqrt (2pi) * δ ^ 3