Square of 11-20

Square of 11-20

11＾2＝121
12＾2＝144
13＾＝169
14＾2＝196
15＾2＝225
16＾2＝256
17＾2＝289
18＾2＝324
19＾1＝361
I hope you can learn it by heart

Who can give me the square of 11 to 20

121 144 169 196 225 256 289 324 361 400

How to calculate the fastest square of a number

Remember the square of some commonly used numbers, and you can certainly improve the speed of doing questions when you are proficient. The square formula of 11 ~ 19 is as follows:
11^2=121
12^2=144
13^2=169
14^2=196
15^2=225
16^2=256
17^289
18^2=343
19^2=361
Then there is the square of a number with bit 5, (A5) ^ 2 = (a) (a + 1) 25
That is 15 ^ 2 = 225
25^2=625
35^2=1225
45^2=2025
……

Li Lin read 5 / 9 of a book in 7 days. What is the average daily reading rate of this book? According to this calculation, how many days can he finish reading this book?

On average, 5 out of 63 people read this book every day and can finish it in 12.6 days
5/9÷7=5/63
So on average, 5 out of 63 people read this book every day
1÷5/63=63/5=12.6
So 12.6 days

A (6-5a, 2a-1) if point a is in the second quadrant, find the value range of A
Tell me why,

Because the abscissa of the point in the second quadrant is less than 0 and the ordinate is greater than 0
So 6-5a0
The solution is a > 6 / 5

How many milliseconds is one second

One millisecond is equal to one thousandth of a second (10-3 seconds). 0.000 000 001 millisecond = 1 picosecond. 0.000 001 millisecond = 1 nanosecond. 0.001 millisecond = 1 microsecond. 1000 millisecond = 1 second

If the equation x ^ 2 / 9-m + y ^ 2 / 4-m = 1 represents an ellipse, find the value range of M

For ellipses, the denominators of X & sup2; and Y & sup2; are positive
So 9-m > 0, 4-m > 0
So m

Given the function f (x) = 1 / 2cos ^ 2x + √ 3 / 2sinxcosx + 1, X belongs to R, (1) find the minimum positive period of function f (x)

Using double angle formula to transform: 1 / 2cos ^ 2x = 1 / 2 * (1 + cos2x) * 1 / 2 = 1 / 4 + 1 / 4cos2x
√3/2sinxcosx=√3/4six2x
So f (x) = √ 3 / 4six2x + 1 / 4cos2x + 5 / 4
Using the auxiliary angle formula, f (x) = 1 / 2 * (√ 3 / 2six2x + 1 / 2cos2x) + 5 / 4
=1/2*sin(2x+π/6)+5/4
So omega = 2
T=2π/ω=π
So the minimum positive period is π

Such as the title
When k is the value, the intersection point of the line x + 2Y + K + 1 = 0, 2x + y + 2K = 0 is in the third quadrant

x+2y+k+1=0,
2x+y+2k=0
x=1/3-k
y=-2/3
The intersection is in the third quadrant, x = 1 / 3-k

Given that f (x) = 3x & # 178; - 2x-1, when f (x) = 0, x =?
I also have two answers, but it seems that the function has only one answer,..... I don't know. Help me
Function is not only one answer, seek explanation

f(x)=3x²-2x-1=0
(3x+1)(x-1)=0
x=-1/3,x=1