Text translation of unit 1 in Volume 2 of grade 6

Text translation of unit 1 in Volume 2 of grade 6

Give me the text

What are the English words in unit 1 of Volume 2 of grade 6

Unit 9 My Story
word:
Attend school
Software
Development
Basic basic
Personal
Leader leadership
Knock
Quickly
Nuts, nuts
Claw claw
Lock lock
This is a new starting point of English teaching materials, sixth grade in the second semester

X + y = 14 5x + 4Y = 60 to find the solution of "X"

6

lgalgb

∵[lga + lgb]^2-4lgalgb
=(lga)^2+(lgb)^2+2lgalgb-4lgalgb
=(lga)^2+(lgb)^2-2lgalgb
=(lga-lgb)^2>=0
∴[lga + lgb]^2>=4lgalgb
Lgalgb

Find the range of the function y = x ^ - ax + 1 (a is a constant), X ∈ [- 1,1]?

The original formula is as follows:
y=(x-a/2)^2+1-a^2/4
This is a parabola with vertex at (A / 2,1-a ^ 2 / 4) and the same shape as y = x ^ 2
Discussion:
When 0 < A / 2 < 1, that is, when 0 < a < 2
The value range is: [1-A ^ 2 / 4,2 + a]
When - 1 < A / 2 < 0, that is - 2 < a < 0
The value range is: [1-A ^ 2 / 4,2-a]
When a / 2 ≤ - 1, that is, when a ≤ - 2
The range is: [2 + A, 2-A]
When a / 2 ≥ 1, that is, when a ≥ 2
The range is: [2-A, 2 + a]
You can get the picture of the above situations

As shown in the figure, the vertex coordinates of the parabola are (5 / 2, - 9 / 8) and pass through point a (8,14). Find the analytic expression of this function

Y = ax ^ 2 + BX + C, take (8,14) in, get 64a + 8b + C = 14
From the vertex we can get - B / 2A = 5 / 2 and B = - 5A
4c-b ^ 2 / a = - 9 / 2 is obtained from (4ac-b ^ 2) / 4A = - 9 / 8
When formula 2 is brought into formula 3, 4c-25a = - 9 / 2
If we take 2 into 1, 24a + C = 14
If 121a = 121 / 2, a = 1 / 2
b=-5/2 c=2
So: y = 1 / 2x ^ 2-5 / 2x + 2

The solution ratio is 1 / 2: 2 / 7 = 1 / 3: (4-x). How to calculate the process
Urgently, there is another 0.8:15 / 4 = (5-x): 0.75

1/2 ：2/7 =1/3 :(4-X )
1/2 ×(4-X ) =2/7 ×1/3
2- (1/2)X =2/21
X =(2- 2/21)÷(1/2) =80/21
0.8：15/4 =（5-X）：0.75
0.8×0.75 =15/4 ×（5-X）
0.6 =75/4 -（15/4）X
X=121/25

What is the result of 100 × 2x + 10 × 2 + X + X

It's 222x. There's no need to divide

Two univariate quadratic equations x ^ - x + M = 0, (m-2) x ^ + 2x-1 = 0 are given
If at least one of the equations has real roots, then the range of M is

Because X & # 178; - x + M = 0 and (m-2) x & # 178; + 2x-1 = 0, at least one equation has a solution. 1. From X & # 178; - x + M = 0, we can get: △ = 1-4m ≥ 0, so: m ≤ 1.2 of 4. From (m-2) x & # 178; + 2x-1 = 0, we can get: △ = 4 + m-2 ≥ 0, so: m ≥ - 2. So when m ≥ - 2 or m ≤ 1 of 4, at least one equation has a real root. So the value range of M is: m ≥ - 2 or m ≤ 1 of 4

If (A-3) & sup2; + | 2b-1 | + | C-4 | = 0, find a + 4b-2c

（a－3）2＋|2b-1|+|c-4|=0
They are all nonnegative numbers, so A-3 = 0 → a = 3, B = 0.5, C = 4
So a + 4b-2c
=3+2-8
=-3