5 A and 3 B, more than 3 times of X, the number of C, 18 minus 6 times of C, 4.2 plus a, and then divided by 0.8

5 A and 3 B, more than 3 times of X, the number of C, 18 minus 6 times of C, 4.2 plus a, and then divided by 0.8

5A+3B 3X+C 18-6C (4.2+A)/0.8

Use the formula containing letters to express the following questions 9 minus 3 times of a, 5 times of X and 12 and 6 times of X and y, 5 times of X is more than 3 times of 0.4, how much less than 3.5 times of X
The sum of one tenth of 9 and the square of a

9 minus 3 times of a
9-3a
The sum of 5 x's and 12's
5x+12
Six times the sum of X and Y
6(x+y)
How much more is 5 times x than 3 0.4
5x-3×0.4
Less than x times 3.5
How much less?
Suppose less a
5 x-a
The sum of one tenth of 9 and the square of a
9 × 1 / 10 + a square

Use the formula containing letters to express the following quantitative relationship. 3.7 more than B______ The sum of 18 A's______ The quotient of X divided by 20______ ; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; is 11.2 more than 5 times X______ ．

The number 3.7 more than B: B + 3.7; the sum of 18 A: 18a; the quotient of X divided by 20: X △ 20; the number 11.2 more than 5 times X: 5x + 11.2; so the answer is: B + 3.7, 18a, X △ 20, 5x + 11.2

The area of a square is 16 cm2. When the side length is increased by X & nbsp; cm, the square area is Y & nbsp; cm2, then the function of Y with respect to X is______ ．

If the side length of the new square is x + 4, then the area y = (4 + x) 2 (x > 0)

88 × (7 × 11-3 / 8) simple operation 12 / 13 × 3 / 7 + 4 / 7 × 12 / 13-12 / 13 simple operation
Kneel down for an answer

88×(7×11-3/8)
=88×7×11-88×3/8
=6776-33
=6743
12/13×3/7+4/7×12/13-12/13
=12/13×(3/7+4/7-1)
=12/13×0
=0

Given the function f (x) = x + 1 / x, X ∈ [1 / 2, a], find the range of function f (x)

f(x)=x+1/x,
Because x ∈ [1 / 2, a], x + 1 / X ≥ 2, if and only if x = 1 / x, take "=",
That is, when x = 1, there is a minimum value of 2
(1) When 1 / 2

There is a rectangular grassland in the middle of a primary school, which is 18 meters long and 10 meters wide. There are two even paths with a width of 2 meters in the middle. How many square meters is the actual area of the grassland
The paths intersect

.（18×10）－（18×2＋10×2－2×2）
=180－52
=128 (M2)
A: the actual area of grassland is 128 square meters

Can 12.5 × 0.56 × 0.4 × 0.8 × 250 be easily calculated

Yes
12.5×0.56×0.4×0.8×250
=（12.5×0.8）×（250×0.4）×0.56
=10×100×0.56
=1000×0.56
=560

Find the general term formula of sequence, A1 = 1, A2 = 2, an = 1 / 3 [a (n-1) + 2A (n-2)]

From an = 1 / 3 [a (n-1) + 2A (n-2)]
Let an + K1 * a (n-1) = K2 * [a (n-1) + K1 * 2A (n-2)]
therefore
k2-k1=1/3,
k2*K1=2/3
The solution is K2 = 1, K1 = 2 / 3 or K2 = - 2 / 3, K1 = - 1
Let K2 = - 2 / 3, K1 = - 1
an-a(n-1)=-2/3[a(n-1)-a(n-2)]
So when n > 2
an-a(n-1)=-2/3[a(n-1)-a(n-2)]=...
=(-2/3)^(n-2)[a(2)-a(1)]
=(-2/3)^(n-2)
When n = 2, an-a (n-1) = (- 2 / 3) ^ (n-2) = 1 also holds
When n > 1
an=an-a(n-1)+.+a2-a1+a1
=a1+a2-a1+.+an-a(n-1)
=a1+(-2/3)^(2-2）+.+(-2/3)^(n-2）
=a1+(-2/3)^(2-2）*[1-(-2/3)^(n-1）]/[1-(-2/3)]
=a1+3/5*[1-(-2/3)^(n-1）]
=8/5-3/5*(-2/3)^(n-1)
When n = 1, an = 8 / 5-3 / 5 * (- 2 / 3) ^ (n-1) = 1 holds
Therefore, the general formula is an = 8 / 5-3 / 5 * (- 2 / 3) ^ (n-1)

A house needs to be paved with square bricks. It needs 128 square bricks with a side length of 3 decimeters. If it needs to use square bricks with a side length of 4 decimeters, how many are needed?
It's going to be solved by equations

You need x blocks
3×3×128=（4×4）x
X = 72