Xiaojun walks about 72 meters per minute. It takes him about 6 minutes to walk from his home to the library. How far is his home from the library?

Xiaojun walks about 72 meters per minute. It takes him about 6 minutes to walk from his home to the library. How far is his home from the library?

72 times 6 equals 432 (m)

Xiaoliang and Xiaojun start from home at 7:20 in the morning and go to the same school. Xiaoliang walks 80 meters every minute and Xiaojun walks 50 meters every minute
Xiaoliang and Xiaojun started from home at 7:20 in the morning and went to the same school. Xiaoliang walked 80 meters every minute and Xiaojun walked 50 meters every minute. Five minutes after Xiaoliang arrived at school, he found that he had left his English book at home and returned immediately. On the way, he met Xiaojun. It was 7:40. How many meters was it from school to their home?
Don't use equations

When we met, we just walked two schools to get home. Xiaojun walked 20 minutes, Xiaoliang walked 20-5 = 15 minutes
[50*20+80*(20-5)]/2
=[1000+1200]/2
=2200/2
=1100
A: it's 1100 meters from home to school
Please accept! Come on

As shown in the picture, Xiaodong starts from home, 65 meters per minute, t minutes to school; Xiaojun starts from home, 60 meters per minute, t minutes to school
Whose home is farthest from the school? How many meters

Distance from Xiaodong's home to school: 65t
Distance from Xiaojun's home to school: 60t
65t-60t=5t
Xiaodong's home is 5T meters away from the school

1 / 2 + 2 / 3 + 3 / 4 +... 2005 / 2006 simple calculation

=1-1/2+1-1/3+…… 1-1/2006=2007-(1+1/2+1/3+…… 1/2006)=2007-(ln2006-γ)≈2007-7.603897968521880840282817113051-0.57721566490153286060651209≈1998,8183879863021289979668922264

As shown in the figure, in △ ABC, the linear equation of the height on the BC side is x-2y + 1 = 0, and the linear equation of the bisector of ∠ A is y = 0. If the coordinates of point B are (1,2), the coordinates of point a and point C are obtained

Point a is the intersection of two lines y = 0 and x-2y + 1 = 0, the coordinates of point a are (- 1, 0).. KAB = 2 − 01 − (− 1) = 1. The equation of bisector line of ∵ - A is y = 0, ∵ - KAC = - 1. The equation of line AC is y = - X-1. BC is perpendicular to x-2y + 1 = 0, ∵ KBC = - 2. The equation of line BC is Y-2 = - 2 (x-1). From y = - X-1, y = - 2x + 4, the coordinates of point a and point C are obtained They are (- 1,0) and (5, - 6) respectively

Using dichotomy to find the approximate value of equation x + 1 / x-3 = 0 in the interval (2,3)

f(x)=x+1/x-3
f(2)=-0.5 f(3)=1/3
F (x) is a decreasing function
x1=2 x2=3
x1\x09 f((x1+x2)/2)\x09 x2\x09 (x1+x2)/2
2.0000000000 \x09-0.1000000000 \x093.0000000000 \x092.5000000000
x1= (x1+x2)/2
2.5000000000 \x090.1136363636 \x093.0000000000 \x092.7500000000
x2=(x1+x2)/2
2.5000000000 \x090.0059523810 \x092.7500000000 \x092.6250000000
2.5000000000 \x09-0.0472560976 \x092.6250000000 \x092.5625000000
2.5625000000 \x09-0.0207078313 \x092.6250000000 \x092.5937500000
2.5937500000 \x09-0.0073914671 \x092.6250000000 \x092.6093750000
2.6093750000 \x09-0.0007229478 \x092.6250000000 \x092.6171875000
2.6171875000 \x090.0026138692 \x092.6250000000 \x092.6210937500
2.6171875000 \x090.0009452484 \x092.6210937500 \x092.6191406250
2.6171875000 \x090.0001110972 \x092.6191406250 \x092.6181640625
2.6171875000 \x09-0.0003059386 \x092.6181640625 \x092.6176757813
two point six one seven one eight seven five zero zero zero

How to take the free unknowns for the basic solution system and the special solution of linear algebraic linear equations?

The fundamental solution system is generally based on the unit basis (1,0,...) ,0）,(0,1,…… 0）,……
The free unknowns of a particular solution are all zero

Square of X + 5x-3 = 0

Square of X + 5x-3 = 0
x²+5x+（5/2）²=3+25/4
（x+5/2）²=37/4
X + 5 / 2 = ± 2 √ 37
X = 2 / 2 (√ 37-5) or x = 2 / 2 (√ 37-5)

Let a = {x | x2 + 4x = 0}, B = {x | x2 + 2 (a + 1) x + A2-1 = 0}, where x ∈ R, if B ⊆ a, find the value range of real number a

A  x | x2 + 4x = 0} = {0, - 4}, ∵ B ⊆ a. ① if B = ∞, △ = 4 (a + 1) 2-4 (A2-1) ＜ 0, a ＜ - 1 is obtained; ② if B = {0}, △ = 0a2 − 1 = 0, a = - 1 is obtained; ③ when B = {- 4}, △ = 0 (− 4) 2 − 8 (a + 1) + A2 − 1 = 0, then the equations have no solution. ④ B = {0, - 4}, − 2 (a + 1)

If the value of the algebraic expression - 3 times the square of X + MX + n times the square of X - x + 10 has nothing to do with the value of X, find the value of M n

It has nothing to do with the value of X, that is, the coefficients of polynomials about X are all 0
-3X²+mX+n+nX²-X+10=(n-3)X²+(m-1)X+10
So n-3 = 0, M-1 = 0
n=3 m=1